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In an arithmetic sequence the common difference is the constant value obtained by subtracting any term from the next term. Here 11 − 7 = 4 and 15 − 11 = 4, so the common difference is 4.
The first term is a, and to reach the nth term we add the common difference exactly (n−1) times. Hence a_n = a + (n−1)d; using n instead of n−1 would give the (n+1)th term.
Here a = 3 and d = 4. Using a_n = a + (n−1)d gives a_10 = 3 + 9×4 = 3 + 36 = 39.
An arithmetic sequence has a constant difference between consecutive terms. In 5, 10, 15, 20 each term increases by 5, while the other lists have changing differences.
Use a_n = a + (n−1)d with a = 5, d = 4 and a_n = 81. Then 81 = 5 + (n−1)4, so (n−1)4 = 76, n−1 = 19, and n = 20.
Here a = 2 and d = 3. The sum formula S_n = n/2[2a + (n−1)d] gives S_20 = 20/2[4 + 19×3] = 10×61 = 610.
The difference between the 9th and 5th terms equals 4 common differences: 33 − 17 = 16. Therefore 4d = 16 and d = 4.
In an arithmetic sequence the middle term is the average of its two neighbours, so 2x = 3 + 11 = 14. Hence x = 7.
The 10th term is S_10 − S_9. S_10 = 100 + 30 = 130 and S_9 = 81 + 27 = 108, so a_10 = 130 − 108 = 22. Equivalently, a_n = S_n − S_{n−1} = 2n + 2.
Using S_n = n/2[2a + (n−1)d], S_10 = 120 gives 2a + 9d = 24, and S_20 = 440 gives 2a + 19d = 44. Subtracting these equations gives 10d = 20, so d = 2.
For an arithmetic sequence, the nth term is a + (n - 1)d. Here a = 4 and d = 5, so solve 4 + (n - 1)5 = 79, giving (n - 1)5 = 75 and n = 16. Thus 79 is the 16th term.
The first n odd numbers form an arithmetic sequence with first term 1 and common difference 2. Their sum is n squared, so for n = 15 the sum is 15 x 15 = 225.
The nth term is a + (n - 1)d. With the 6th term equal to 20 and d = 3, we have 20 = a + 5 x 3, so a = 20 - 15 = 5.
Use S_n = (n/2)[2a + (n - 1)d]. Substituting S_20 = 400, a = 1 and n = 20 gives 400 = 10(2 + 19d), so 40 = 2 + 19d and d = 2.
For any sequence, the nth term is S_n - S_(n-1). Here a_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 6n + 2, which is linear with common difference 6.
The numbers form the arithmetic sequence 10, 13, 16, ..., 97. Here a = 10, d = 3 and 97 = 10 + (n - 1)3, so n = 30. The sum is (30/2)(10 + 97) = 15 x 107 = 1605.
Let the numbers be a - d, a and a + d. Their sum is 3a = 24, so a = 8. Their product is (8 - d)(8)(8 + d) = 440, giving 64 - d^2 = 55 and d = 3, so the numbers are 5, 8 and 11.
Subtracting the 4th term from the 11th term gives 7d = 32 - 11 = 21, so d = 3 and a = 11 - 3 x 3 = 2. Then S_15 = (15/2)[2 x 2 + 14 x 3] = (15/2)(46) = 345.
Let the first term be a and common difference d. From S_10 = 15 and S_15 = 10, we get 2a + 9d = 3 and 2a + 14d = 4/3, so d = -1/3 and a = 3. Then S_25 = (25/2)[2 x 3 + 24 x (-1/3)] = (25/2)(-2) = -25.
Since the 7th term is 5 times the 2nd term, a + 6d = 5(a + d), which gives d = 4a. The 10th term condition a + 9d = 37 then gives 37a = 37, so a = 1 and d = 4. Hence S_12 = (12/2)[2 x 1 + 11 x 4] = 6 x 46 = 276.
The angle in a semicircle is a right angle, so ∠ACB = 90°. The angles of triangle ABC add to 180°, hence ∠BAC = 180° − 90° − 35° = 55°.
A tangent is perpendicular to the radius at the point of contact, so OP is perpendicular to the tangent line. Therefore the perpendicular distance from O to the tangent is exactly the radius, 8 cm.
The radius is perpendicular to the tangent at the point of contact, so ∠OAP = ∠OBP = 90°. In quadrilateral AOBP the angles add to 360°, giving ∠AOB = 360° − 90° − 90° − 60° = 120°.
The perpendicular from the centre to a chord bisects the chord, so each half is 8 cm. This half-chord, the distance from the centre and the radius form a right triangle, giving distance = √(10² − 8²) = 6 cm.
In a cyclic quadrilateral, each pair of opposite angles is supplementary, so ∠A + ∠C = 180°. Therefore ∠C = 180° − 70° = 110°.
By the tangent–chord (alternate segment) theorem, the angle between a tangent and a chord equals the angle made by that chord in the alternate segment. Hence ∠ACB = ∠PAB = 35°.
Since the radius is perpendicular to the tangent at each point of contact, ∠OAP = ∠OBP = 90°. In quadrilateral AOBP, ∠APB = 360° − 90° − 90° − 110° = 70°.
Triangle AOB is right-angled at O with OA = OB = 7 cm. By Pythagoras, AB² = 7² + 7² = 98, so AB = 7√2 cm.
For a quadrilateral circumscribing a circle, the sums of opposite sides are equal: AB + CD = BC + DA. Thus 6 + 5 = 7 + DA, so DA = 4 cm.
Each tangent is perpendicular to the radius at the point of contact, so triangles OAP and OBP are right-angled at A and B. Their common hypotenuse OP is 10 cm and OA = OB = 6 cm, so PA = PB = √(10² − 6²) = 8 cm. Each triangle has area ½ × 6 × 8 = 24 cm², and the quadrilateral has area 48 cm².
The radius drawn to the point of contact is perpendicular to the tangent. Therefore the angle between OP and the tangent at P is a right angle, that is 90°. This tangent–radius property is fundamental in all tangent problems.
A point outside a circle lies outside its boundary, so two distinct tangents can be drawn from it, one on each side. A point on the circle gives exactly one tangent, and an interior point gives none.
In a cyclic quadrilateral, an exterior angle is equal to the interior opposite angle. Here ∠CBE is the exterior angle at B, so it equals the interior angle at the opposite vertex D, that is ∠ADC. Hence ∠ADC = 75°.
The perpendicular from the centre bisects each chord. For the 24 cm chord, half is 12 cm, so its distance from the centre is √(13² − 12²) = 5 cm. For the 10 cm chord, half is 5 cm, so its distance is √(13² − 5²) = 12 cm. Since the chords are on opposite sides, the distance between them is 5 + 12 = 17 cm.
The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circle. Here the central angle is 110°, so the inscribed angle on the major arc is 110°/2 = 55°. The major arc is the part not containing the minor arc's central angle, so the half-angle relation applies directly.
Tangents drawn from the same external point to a circle are equal in length. If the tangent lengths from A, B and C are x, y and z, then x + y = 10, y + z = 12 and z + x = 8. Adding the first and third equations and subtracting the second gives 2x = 6, so x = 3 cm.
For two circles touching externally, the distance between centres is 5 + 3 = 8 cm. A direct common tangent forms a right triangle with the difference of radii, so its length is √(8² − (5 − 3)²) = √(64 − 4) = √60 cm, which equals 2√15 cm.
In quadrilateral OAPB, the radii OA and OB are perpendicular to the tangents PA and PB, so ∠OAP = ∠OBP = 90°. Hence ∠APB + ∠AOB = 180°. With the ratio 1 : 3, let ∠APB = x and ∠AOB = 3x; then 4x = 180°, so x = 45°.
Probability is measured on a scale from 0 to 1. A sure event has probability 1 because it happens in every trial, while an impossible event has no favourable outcome at all, so the ratio of favourable outcomes to total outcomes is 0. Hence the probability of an impossible event is 0.
The sample space is {1, 2, 3, 4, 5, 6}, which gives six equally likely outcomes. The prime faces are 2, 3 and 5, that is, three favourable outcomes. So the probability is 3/6, which simplifies to 1/2.
The total number of equally likely outcomes is the total number of balls, 3 + 5 = 8. The favourable outcomes are the 3 red balls. Therefore the probability of drawing a red ball is 3/8.
The sample space is {HH, HT, TH, TT}, giving four equally likely outcomes. The event 'at least one head' contains HH, HT and TH, because only TT has no head. So the probability is 3/4, which also equals 1 minus the probability 1/4 of getting no head.
A pack has 12 face cards, of which 6 are red: the jack, queen and king of hearts and the jack, queen and king of diamonds. So there are 6 favourable outcomes out of 52. The probability is 6/52, which simplifies to 3/26.
Rolling two dice gives 6 x 6 = 36 equally likely ordered outcomes. A total of 9 arises from (3, 6), (4, 5), (5, 4) and (6, 3), which is four favourable outcomes. The probability is therefore 4/36 = 1/9.
The sectors that are not red are the 2 blue and 5 green sectors, that is, 7 of the 10 equal sectors. So the probability is 7/10. Equivalently, it is 1 minus the probability 3/10 of stopping on red.
If the total number of balls is n, then the probability of drawing red is 5/n. Setting 5/n = 1/3 gives n = 15, so the bag holds 15 balls altogether. Of these, 15 - 5 = 10 are blue, and indeed 5/15 = 1/3.
Three coins give 2 x 2 x 2 = 8 equally likely outcomes. Exactly two heads occur in HHT, HTH and THH, which is three favourable outcomes, while HHH and the outcomes with one or no head are excluded. Hence the probability is 3/8.
There are 36 equally likely ordered outcomes. The prime totals possible are 2, 3, 5, 7 and 11, occurring in 1, 2, 4, 6 and 2 outcomes respectively, which gives 1 + 2 + 4 + 6 + 2 = 15 favourable outcomes. So the probability is 15/36 = 5/12.
Probability is measured on a scale from 0 to 1. An event that is certain to occur includes all possible outcomes, so its probability is 1, while 0 is reserved for an impossible event.
The word MATHEMATICS has 11 letters in total. Its vowels are A, E, A and I, giving 4 vowel slips, so the probability is 4/11.