Question

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A. P. is 43, find the nth term.

Answer 1

$T_{17}$ = 5 + 2a₈

Given that $T_{11}$ = 43

W know that $T_n$ = a + (n-1) d ( where a = first term and d = common difference of A.P)

Hence, equating we have
a + (17 - 1) d = 5 + 2{ a + (8 - 1 ) d}

a + 16 d = 5 + 2a + 14d

or 2d - 1a = 5 .......(1)


Also, 1a + (11 - 1)d = 43

Or 1 a + 10d = 43 .........(2)


Solving 1 and 2 we have

12d = 48

d = 4

Putting the value of d= 4 in

1 a + 10d = 43

or 1a + 10* 4=43

or 1a + 40=43

or 1a =43 -40

or a = 3

Hence $T_n$ = 3 + (n-1)4

= 3 +4n -4

or $T_n$ = 4n -1