Question

The sum of fourth and the ninth terms of an AP is 46 and their product is 465. Find the sum of first 10 terms of this AP

Answer 1

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Answer 2

Answer:

Formula of nth term of AP = $a_n=a+(n-1)d$

Substitute n = 4

$a_4=a+(4-1)d$

$a_4=a+3d$

Substitute n = 9

$a_9=a+(9-1)d$

$a_9=a+8d$

We are given that The sum of fourth and the ninth terms of an AP is 46

So,$a+3d+a+8d=46$

$2a+11d=46$

$a=\frac{46-11d}{2}$  ---1

We are also given that the product of 4th and 9th term is 465

$(a+3d)(a+8d)=465$

$a(a+8d)+3d(a+8d)=465$

$a^2+8ad+3ad+24d^2=465$

$a^2+11ad+24d^2=465$

Substitute the value of a from 1

$(\frac{46-11d}{2})^2+11(\frac{46-11d}{2})d+24d^2=465$

$d = \frac{16}{5},\frac{-16}{5}$

When d = 16/5

$a=\frac{46-11(\frac{16}{5})}{2}$

$a=5.4$

So, sum of first n terms = $S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 10

$S_{10}=\frac{10}{2}(2(5.4)+(10-1)(\frac{16}{5}))$

$S_{10}=198$

when d = -16/5

$a=\frac{46-11(\frac{-16}{5})}{2}$

$a=40.6$

So, sum of first n terms = $S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 10

$S_{10}=\frac{10}{2}(2(40.6)+(10-1)(\frac{-16}{5}))$

$S_{10}=262$