Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, .....

Question

Let the sum of the first n terms of a non-constant
A.P., a1, a2, a3, ..... be $$50n + {{n(n - 7)} \over 2}A$$, where
A is a constant. If d is the common difference of
this A.P., then the ordered pair (d, a50) is equal to

Accepted Answer

Sn = $$50n + {{n(n - 7)} \over 2}A$$

We know, nth tem

Tn = Sn - Sn - 1

= $$50n + {{n(n - 7)} \over 2}A$$ - $$50\left( {n - 1} \right) - {{\left( {n - 1} \right)\left( {n - 8} \right)} \over 2}A$$

= 50 + $${A \over 2}\left[ {{n^2} - 7n - {n^2} + 9n - 8} \right]$$

= 50 + A(n - 4)

We also know, common difference

d = Tn - Tn - 1

= 50 + A(n - 4) - 50 - A(n - 5)

= A

And T50 = 50 + A(50 - 4)

= 50 + 46A

$$ \therefore $$ (d, a50) = (A, 50+46A)

Solutions · 1