JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 17)
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
(A, 50+45A)
(50, 50+45A)
(A, 50+46A)
(50, 50+46A)
Explanation
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)
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)
Comments (0)
