Question

in an ap the sum of its first 10 terms is -80 and the sum of its next 10 terms is -280 find ap

Answer 1

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Answer 2

Answer:

The formed AP is 1 , -1 , -3 , -5 , . .

Step-by-step explanation:

We know,

Sum of n terms in AP is given as :

$S_{n}$ = $\frac{n}{2}[2a+(n-1)d]$

where,

a =  First term in series

d = common difference

n = Number of terms in AP

Let first term be $a_{1}$

difference be d

Given,

Sum of first 10 terms in AP is -80

Substituting values we get,

- 80 = $\frac{10}{2}[(2a_{1}+(10-1)d]$

-80 = $\frac{10}{2}[(2a_{1}+9d]$

-80 = 10$a_{1}$+45d

10$a_{1}$+45d = -80         [ Let this be Eqn 1 ]

According to question,

The sum of next 10 terms is -280, that is,

$S_{20}-S_{10}$ = -280

$\frac{20}{2}[(2a_{1}+(20-1)d]$ - $\frac{10}{2}[(2a_{1}+(10-1)d]$ = -280

$\frac{20}{2}[(2a_{1}+19d]$] - $\frac{10}{2}[(2a_{1}+9d]$ = -280

20$a_{1}$+190d - 10$a_{1}$-45d = -280

10$a_{1}$+145d = -280      [ Let this be Eqn 2]

Subtracting Eqn 2 from Eqn 1,

10$a_{1}$+145d - ( 10$a_{1}$+45d ) = -280 - ( -80 )

10$a_{1}$+145d - 10$a_{1}$-45d ) = -280 + 80

100d = -200

d = -2

Substituting value of d in Eqn 1,

10$a_{1}$+45(-2) = -80    

10$a_{1}$-90 = -80    

10$a_{1}$ = -80+90

10$a_{1}$ =10

$a_{1}$ = 1

The AP is formed as :

$a_{1}+(a_{1}+d)+(a_{1}+2d)+(a_{1}+3d)+. .$

Substituting values of $a_{1}$ and d ;

1 , (1+(-2)) , (1+2(-2)) , (1+3(-2)) ,  .  .

1 , -1 , -3 , -5 , . .

Hence,

The formed AP is 1 , -1 , -3 , -5 , . .