Question

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 AP.

Answer 1

Step 1: Write the formula for the sum of n terms

`S_n = n/2 [2a + (n-1)d]`

Step 2: Use the information about the sum of the first 9 terms

`S_9 = 9/2 [2a + (9-1)d]`

`153 = 9/2 [2a+8d]`

306 = 9[2a + 8d]

`2a+8d = 306/9 = 34`

2a + 8d = 34

Step 3: Use the information about the sum of the last 6 terms

S_{last 6} = `6/2 [2l - (6-1)d]`

Here, l = a + 39d (the last term). Substitute l:

S_{last 6} = `6/2 [2 (a+39d)-5d]`

687 = 3[2a + 78d − 5d]

687 = 3[2a + 73d]

Divide through by 3:

`2a + 73d=687/3 = 229`

2a + 73d = 229

Step 4: Solve the system of equations

(2a + 73d) − (2a + 8d) = 229 − 34

65d = 195

`d = 195/65 = 3`

2a + 8(3) = 34

2a + 24 = 34

2a = 10

a = 5

a = 5, d = 3

The sum of all 40 terms is:

`S_40 = 40/2 [2a+(40-1)d]`

S₄₀​ = 20 [2(5) + 39(3)]

S₄₀ ​ = 20[10 + 117]

S₄₀  = 20 × 127 = 2540