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 the A.P. Also, find the sum of all the terms of the A.P.

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Step 1: Here $n = 40$.
$S_9 = \frac{9}{2}[2a + 8d] = 153 \Rightarrow a + 4d = 17$ ... (i)
Step 2: and $S_{40} - S_{34} = 687$ or $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$
$\Rightarrow 6a + 219d = 687$ or $2a + 73d = 229$ ... (ii)
Step 3: Solving (i) and (ii) to get $a = 5, d = 3$
Step 4: Also, $S_{40} = \frac{40}{2}(10 + 39 	imes 3) = 2540$

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