Question

how many terms of an ap -10, -7, -4, -1 ........ must be added to get the sum 104​

Answer 1

Answer:

hey mate there's your answer

given series of AP is -10,-7,-4,-1..................

let first term of the AP be a1= -10

second term of the AP be a2=-7

and common difference be d.

d=a2-a1

d=-7-(-10)

d=-7+10

d=3

so common difference is d=3

we know that,

Sum of n terms in the AP is,

Sn=n/2[2a1+(n-1)d]

104=n/2[2×-10 +(n-1)3]

104×2=n[-20+3n-3]

208=n[3n-23]

208=3n^2-23n

3n^3-23n-208=0

3n^2-39n+16n-208=0

3n(n-13)+16(n-13)=0

(3n+16)(n-13)=0

3n+16=0 and n-13=0

n=-16/3 n=13

n can't be a fraction so n=13

so sum of 13 terms in the AP is 104.

hope it helps you...............

thank you!!!

Answer 2

Answer:

ap :- -10 , -7, -4 , -1

a= 10

d=-7-(-10) = -7+10=3

sn = 104

n/2 {2a+(n-1)d}=104

n{2(-10)+(n-1)3}=208

n{-20+3n-3}=208

n(3n-23) =208

3n^2-39n-208=0

3n^2-39n+16n-208=0

3n(n-13) + 16(n-13)=0

(3n+16) (n-13)=0

3n+16=0 , n-13=0

3n= -16 , n= 13

n= -16/3 , n = 13

n cannot be negative or in fraction .

therefore , 13 terms of the ap must be added to get the sum 104 .