Question

The sum of 5th and 9th terms of an AP is 8 and their product is 15.find d sum of first 28 terms of AP

Answer 1

a+4d+a+8d=8
⇒2a=8-12d
∴a=4-6d  

(a+4d)(a+8d)=15
⇒(4-6d+4d)(4-6d+8d)=15
⇒(4-2d)(4+2d)=15
⇒16-4d²=15
∴d=1/2

∴a=4-6d=1

S(28)=n/2(a+(n-1)d)=14(1+27/2)=14*29/2=7*29=203

i hope it helps!!

Answer 2

Answer:

The sum of first 28 terms can be 217 or 7

Step-by-step explanation:

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

Substitute n = 5

$a_5=a+(5-1)d$

$a_5=a+4d$

Substitute n = 9

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

$a_9=a+8d$

We are given that The sum of 5th and 9th terms of an AP is 8

So, $a_5+a_9=8$

$a+4d+a+8d=8$

$2a+12d=8$

$a+6d=4$

$a=4-6d$ ----1

We are given that the product of 5th and 9th term is 15

$(a+4d)(a+8d)=15$

$a^2+8ad+4ad+32d^2=15$

$a^2+12ad+32d^2=15$  ---2

Substitute the value of a from 1 in 2

$(4-6d)^2+12(4-6d)d+32d^2=15$

$d=\frac{1}{2},\frac{-1}{2}$

At d = 1/2

$a=4-6\frac{1}{2}=1$

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

Substitute n = 28

$S_{28}=\frac{28}{2}(2(1)+(28-1)\frac{1}{2})$

$S_{28}=217$

At d = -1/2

$a=4-6\frac{-1}{2}=7$

$S_{28}=\frac{28}{2}(2(7)+(28-1)\frac{-1}{2})$

$S_{28}=7$

Hence The sum of first 28 terms can be 217 or 7