Question

35. in an A.P if the sum of the third term and seventh term is 6 and their product is 8, then find the
sum of first 16 terms.
​

Answer 1

AnswEr:

We know the Formula:

$\dag$$\large\boxed{\sf{\red{a_n = a + (n - 1)d}}}$

So, 7th term =$\sf\;a_7 = a + 6d$

And, 3rd term = $\sf\;a_3 = a + 2d$

According to Question:

Sum of the Third and Seventh Term of the AP.

$\dag\sf\; a_3 + a_7 = 6$

$:\implies\sf\; a + 2d + a + 6d = 6$

$:\implies\sf\; 2a + 8d = 6$

$:\implies\sf\; 2(a + 4d) = 6$

$:\implies\sf\; a + 4d = \dfrac{6}{2}$

$:\implies\sf\; a + 4d = 3$

$:\implies\sf\;a = 3 - 4d$ __eq(1)

Now, Product of Third and Seventh Term

$\dag\sf\; a_3 \times\;7 = 8$

$:\implies\sf\; (a + 2d) (a + 6d) = 8$

$:\implies\sf\; a(a + 6d) + 2d (a + 6d) = 8$

$:\implies\sf\; (a + 2d) (a + 6d) = 8$ _eq(1)

$\rule{150}3$

$\dag$ From Equation (1)

$:\implies\sf\; (3d - 2d) (a + 6d) = 8$

$:\implies\sf\; 9 - 4d^2 = 8$

$:\implies\sf\; d = \dfrac{1}{2}$

$\rule{150}3$

$\dag$ $\sf\; If \; d = \dfrac{ -1}{2}$

$:\implies\sf\; S_16 = \dfrac{16}{2}(2 + \frac{15}{2})$

$:\implies\sf\; S_16 =8 \times\; \dfrac{19}{2}$

$:\implies\large{\underline{\boxed{\sf{\blue{x = 76}}}}}$

$\dag$$\sf\;If \; d =\dfrac{1}{2}$

$:\implies\sf\;S_16 = \dfrac{16}{2}( 10 - \frac{15}{2})$

$:\implies\sf\;S_16 = 8 \times\; \dfrac{5}{2}$

$:\implies\large{\underline{\boxed{\sf{\blue{x = 20}}}}}$

Therefore, If $\sf\; d = \dfrac{-1}{\;2}$ the sum of the first sixteen terms of the AP is 76.

If If $\sf\;d = \dfrac{ 1}{2}$ the sum of the first sixteen terms of the AP is 20.

$\rule{150}3$