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.

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Solution

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.
Let the first term and common difference of A.P. be a and d.

$∵$ Sum of n terms of A.P., $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$S_{9} = \frac{9}{2} [2 a + (9 - 1) d] = \frac{9}{2} [2 a + 8 d] = 9 (a + 4 d)$

$\Rightarrow 153 = 9 (a + 4 d)$ [ $∵$ Given, S9 = 153]

$\Rightarrow 17 = a + 4 d$

$\Rightarrow a + 4 d = 17$ ...(i)

Similarly, sum of 6 terms, $S_{6} = \frac{6}{2} [2 a + (6 - 1) d]$

$\Rightarrow 687 = 3 (2 a + 5 d)$

$\Rightarrow 229 = 2 a + 5 d$

$\Rightarrow 2 a + 5 d = 229$ ...(ii)

On solving eqs (i) & (ii), we get $3 d = - 195 \Rightarrow d = - 65$

Substituting value of d in eq (i), we get $a + 4 (- 65) = 17$

$\Rightarrow a = 17 + 260 = 277$

Thus, first term of given A.P. is 277 and common difference is –65.

Now, Sum of all terms i.e.,

$S_{40} = \frac{40}{2} [2 \times 277 + (40 - 1) (- 65)]$

$= 20 [554 + 39 (- 65)]$

$= 20 (554 - 2535)$

$= 20 \times (- 1981) = - 39, 620$

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