Q) In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.
Ans:
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Let’s consider the first terms of AP is a and the common difference is d.
(i) Calculating first term and common difference of AP:
Since the sum of first n terms of an AP,
∴ Sum of first 9 terms, =
[2a + (9 – 1) (d)] = 9 (a + 4 d)
Since it is given that the sum of first 9 terms is 153
∴ 153 = 9 (a + 4 d)
∴ 17 = a + 4d
∴ a + 4 d = 17 …… (i)
Next, it is given that the sum of last 6 terms is 687
Since, Sum of last 6 terms = Sum of first 40 terms – Sum of first 34 terms
∴ = 687
∴ [2a + (40 – 1) (d)] –
[2a + (34 – 1) (d)] = 687
∴ 20 (2 a + 39 d) – 17 (2 a + 33 d) = 687
∴ 40 a + 780 d – 34 a – 561 d = 687
∴ 6 a + 219 d = 687
∴ 2 a + 73 d = 229 ….. (ii)
By multiplying equation (i) by 2 and then subtracting equation (ii), we get:
2 (a + 4 d) – (2 a + 73 d) = 2 x 17 – 229
∴ 2 a + 8 d – 2 a – 73 d = 34 – 229
∴ – 65 d = – 195
∴ d = 3
By substituting value of d in equation (i), we get:
a + 4d = 17
∴ a + 4 (3) = 17
∴ a + 12 = 17
∴ a = 17 – 12
∴ a = 5
Therefore, first term of given AP is 5 and common difference of AP is 3.
(ii). Sum of all terms of AP:
Since the sum of n terms of an AP,
It is given that AP has 40 terms, a = 5, d = 3
∴ Sum of 40 terms, [2 (5) + (40 – 1) (3)]
∴ = 20 (10 + (39) (3)] = 20 (10 + 117) = 20 x 127 = 2,540
Therefore sum of its 40 terms is 2,540
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