Question

If the first four terms of an arithmetic progression are p , p + 2q , 3p + q and 30 . find 2016th term

Answer 1

∴$a_{2016} =16126$

Step-by-step explanation:

Given,

The first four terms of an arithmetic progression are p , p + 2q , 3p + q and 30.

∴ $p+2q-p=3p+q-p-2q=30-3p-q$ = Common difference

⇒$p+2q-p=3p+q-p-2q or 3q =2p$   ....(1)

Also,

$3p+q-p-2q=30-3p-q$

⇒$5p=30$

⇒$p=d\frac{30}{5} =6$

Put p =6 in (1), we get

$3q =2\times 6=12$

∴$q =4$

The first four terms of an arithmetic progression are 6 , 6 + 2(4) , 3(6) + 4 and 30.

i.e., 6, 14, 22 and 30

Here, first term(a) =, common difference(d) = 8, n = 2016

∴ $a_{2016} =6+(2016-1)\times 8$

= 6 + 16120

= 16126

Hence, $a_{2016} =16126.$