Question

If the sum of first 7 terms and 15 terms of an a.p are 98 and 390 respectively then find the sum of first 10 terms​

Answer 1

SoluTion :-

$\fbox {\sf Given}$

$\rightarrow \: \sf S_{7}= 98$

It can be also written as

$\sf \to \frac{7}{2} (2a + 6d) = 98 \\ \\ \sf \to 14a + 42d = 196 \\ \\ \sf \to a + 3d = 14 - - - \: eq \: (i)\\\\ \rightarrow \: \sf S_{15}= 390$

It can be also written as

$\sf \to \frac{15}{2} (2a + 14d) = 390 \\ \\ \sf \to 30a + 210d = 780 \\ \\ \to\sf a + 7d = 26 \: - - - \: eq \: (ii)$

Subtract eq (i) from eq (ii)

$\sf \implies a + 7d - ( a + 3d ) = 26 - 14 \\ \\ \sf \implies</p><p>4d = 12 \\ \\ \sf \implies</p><p>d = \frac{12}{4} \\ \\ \sf \implies</p><p>d = 3 </p><p>$

Putting the value of d = 3 in equation (i)

$\sf \implies a + 3(3) = 14 \\ \\ \sf \implies </p><p>a + 9 = 14 \\ \\ \sf \implies</p><p>a = 5</p><p>$

From the above calculation , we can say that , the given AP is 5, 8, 11, 14, 17, 20...

Find the sum of first 12 terms, so,

$\sf S_{12} = \frac{12}{2} \bigg(2 \times 5 + (11)3 \bigg) \\ \\ \sf= 6(10 + 33) \\ \\ \sf= 258$

The sum of first 12 terms of given AP is 258.