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Let the sum of the first n terms of a no...

Let the sum of the first n terms of a non-constant A.P., `a_(1), a_(2), a_(3),... " be " 50n + (n (n -7))/(2)A`, where A is a constant. If d is the common difference of this A.P., then the ordered pair `(d, a_(50))` is equal to

A

(50, 50+46A)

B

(A, 50 + 45 A)

C

(A, 50 + 46A)

D

(50, 50 + 45A)

Text Solution

Verified by Experts

The correct Answer is:
C
`Sn=50n + (n(n-7))/(2) A rArr T_(n)=S_(n)-S_(n-1)`
`rArr T_(n)=50(n-n+1)+A/2 [n(n-7)-(n-1)(n-8)]`
`T_(n)=50 + A/2 [-7n+9n-8] rArr T_(n)=50+A(n-4) ` and
`d=T_(5)-T_(4)=(50+A)-(50)=A rArr a_(50)=50+46A`
`(d,9_(50))=(A,50+46A)`
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