how many terms of an ap must be taken for their sum to be equal to 91 if its third term is 9 and the difference between the seventh and the second term is 20.?
Ans- 7
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Answers
Step-by-step explanation:
there is a need of 7 terms to get the sum =91
Step-by-step explanation:
Given that 3rd term = 9
nth term = a+(n-1)d
Therefore 3rd term = a+(3-1)d
=>> a+2d = 9 -----(1)
Then , difference b/w 7th and 2nd term = 20
=>> a+(7-1)d -[a+(2-1)d] = 20
=>> a+6d - a-d = 20
=>> 5d = 20
=>> d = 4
Substituting [d = 4] in eq.(1)
=>> a+2(4) = 9
=>> a+8 = 9
=>> a = 9-8 = 1
=>> a = 1
Sum of n terms = n/2{2a+(n-1)d}
Sum = 91
n = ?
a = 1
d = 4
=>> 91 = n/2{2(1)+(n-1)4}
=>> 91 × 2 = n{2+4n-4}
=>> 182 = n(4n-2)
cancelling 2 from both sides
=>> 91 = n(2n-1)
=>> 91 = 2n^2 - n
=>> 2n^2 - n - 91 = 0
=>> 2n^2 -14n +13n -91 = 0
=>> 2n(n-7) +13(n-7) = 0
=>>(2n+13)(n-7) = 0
=>>2n+13 = 0 or n-7 = 0
=>> 2n = -13 or n = 7
=>> n = -13/2
Hence n = 7