Let the sum of the first n terms of a non-constant A-P- a1-a2-a3- - be 50n-nn-72A- where A is a constant- If d is the common difference of this A-P- then the orderd pair d-a50 is equal to-

S_n=50n+n(n 7)/2A and we know that nth term, T_n=S_n S_n 1 =50n+n(n 7)2A 50(n 1) (n 1)(n 8)2A =50+A2[n^2 7n n^2+9n 8] =50+A(n 4) and d=T_n T_n 1 =50+A(n 4) ...

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Standard XI

Mathematics

Sum of n Terms

Let the sum o...

Question

Let the sum of the first n terms of a non-constant A.P., $a_{1}, a_{2}, a_{3}, \dots$ be $50 n + \frac{n (n - 7)}{2} A$ , where $A$ is a constant. If $d$ is the common difference of this A.P., then the orderd pair $(d, a_{50}$ ) is equal to:

(50, 50 + 46 A)

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(A, 50 + 46 A)

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(A, 50 + 45 A)

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(50, 50 + 45 A)

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Solution

The correct option is B $(A, 50 + 46 A)$
$S_{n} = 50 n + \frac{n (n - 7)}{2} A$
and we know that
$n t h term, T_{n} = S_{n} - S_{n - 1} = 50 n + \frac{n (n - 7)}{2} A - 50 (n - 1) - \frac{(n - 1) (n - 8)}{2} A = 50 + \frac{A}{2} [n^{2} - 7 n - n^{2} + 9 n - 8] = 50 + A (n - 4)$
$and d = T_{n} - T_{n - 1} = 50 + A (n - 4) - 50 - A (n - 5) = A ∴ T_{50} = 50 + A (50 - 4) ∴ (d, a_{50}) = (A, 50 + 46 A)$

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Let the sum of the first n terms of a non-constant A.P., a1,a2,a3,⋯ be 50n+n(n−7)2A, where A is a constant. If d is the common difference of this A.P., then the orderd pair (d,a50) is equal to:

The correct option is B (A,50+46A)Sn=50n+n(n−7)2A and we know that nth term,Tn=Sn−Sn−1=50n+n(n−7)2A−50(n−1)−(n−1)(n−8)2A=50+A2[n2−7n−n2+9n−8]=50+A(n−4) and d=Tn−Tn−1=50+A(n−4)−50−A(n−5)=A∴T50=50+A(50−4)∴(d,a50)=(A,50+46A)

Let the sum of the first n terms of a non-constant A.P., $a_{1}, a_{2}, a_{3}, \dots$ be $50 n + \frac{n (n - 7)}{2} A$ , where $A$ is a constant. If $d$ is the common difference of this A.P., then the orderd pair $(d, a_{50}$ ) is equal to: