**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:**

**VIDEO SOLUTION**

**STEP BY STEP SOLUTION**

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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