Question

Find the sum of first 20 terms of an A.P., in which 3rd term is 8 and 7th term is 9 more than its 4th term. ​

Answer 1

It is given in the question :

• 3rd term of AP is 8

• 7th term is 9 more than its 4th term

${\underline{\frak{We\;have}}} \begin{cases} & \sf a_3 = 8 \\ & \sf a_7 = a_4 + 9 \end{cases}$

It is asked to find out , the sum of 20 terms of an A.P.

So, we know :

• a + 2d = 8 ...(1)

• a + 6d = a + 3d + 9 ..(2)

$\sf a_3 = 8$

$\: \: :\implies\sf a + 6d = a + 3d + 9$

$\: \: :\implies\sf \cancel{a} + 6d = \cancel{a} + 3d + 9$

$\: \: :\implies\sf 6d - 3d = 9$

$:\implies\sf 3d = 9$

$:\implies\sf d = \dfrac{9}{3}$

$:\implies\bf d = 3$

Putting in eq (1)

$:\implies\sf a + 2(3) = 8$

$:\implies\sf a + 6 = 8$

$:\implies\sf a = 8 - 6$

$:\implies\bf a = 2$

Now, sum of 20th term of A.P is :

${\boxed{\sf{S_n = a + \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}}}$

So ,

Putting

• a = 2

• d = 3

• n = 20

$\sf {S_n = a + \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}$

$\sf {S_n = 2 + \dfrac{20}{2} \bigg\lgroup\sf 2(2) + (20 - 1)(3) \bigg\rgroup}$

$\sf {S_n = 2 + 10 \bigg\lgroup\sf 4 + (19)(3) \bigg\rgroup}$

$\sf {S_n = 2 + 10 \bigg\lgroup\sf 4 + 57 \bigg\rgroup}$

$\sf {S_n = 2 + 10 \bigg\lgroup\sf 61 \bigg\rgroup}$

$\sf {S_n = 2 + 610 }$

$\sf {S_n = 612 }$

Hence,

${\boxed{\sf{Sum \: of \: 20 \: Terms = {\red{612}}}}}$