JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 14)
Let a1, a2, ..., an be a given A.P. whose
common difference is an integer and
Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $$ \le $$ n $$ \le $$ 50, then
the ordered pair (Sn-4, an–4) is equal to:
common difference is an integer and
Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $$ \le $$ n $$ \le $$ 50, then
the ordered pair (Sn-4, an–4) is equal to:
(2480, 249)
(2480, 248)
(2490, 248)
(2490, 249)
Explanation
$${a_n} = {a_1} + (n - 1)d$$
$$ \Rightarrow 300 = 1 + (n - 1)d$$
$$ \Rightarrow (n - 1)d = 299 = 13 \times 23$$
since, n $$ \in $$[15, 50]
$$ \therefore $$ n = 24 and d = 13
$${a_{n - 4}} = {a_{20}} = 1 + 19 \times 13 = 248$$
$$ \Rightarrow {a_{n - 4}} = 248$$
$${S_{n - 4}} = {{20} \over 2}\{ 1 + 248\} = 2490$$
$$ \Rightarrow 300 = 1 + (n - 1)d$$
$$ \Rightarrow (n - 1)d = 299 = 13 \times 23$$
since, n $$ \in $$[15, 50]
$$ \therefore $$ n = 24 and d = 13
$${a_{n - 4}} = {a_{20}} = 1 + 19 \times 13 = 248$$
$$ \Rightarrow {a_{n - 4}} = 248$$
$${S_{n - 4}} = {{20} \over 2}\{ 1 + 248\} = 2490$$
Comments (0)
