Question

The 7th term from the end of the AP - 11, - 8 ,- 5, ... , 49 is

Answer 1

Answer:

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Then the 29th term.

Answer 2

Given:

AP - 11, - 8 ,- 5, ... , 49

To Find:

The 7th term from the end of the AP

Solution:

An arithmetic progression is a progression in which every consecutive term differs by a common difference which is denoted by d and the first term of an AP is denoted by a.

The nth term of an AP is given by,

$T_n=a+(n-1)d$

The given AP is - 11, - 8 ,- 5, ... , 49

where

first term a=-11

common difference=-8+11

                               =3

Now to find the last term number which will be,

$T_n=a+(n-1)d\\49=-11+(n-1)3\\3(n-1)=60\\n-1=20\\n=21$

So the number for the 7th term from the end would be,

n=21-7+1

 =15th

So the value of the 15th will be,

$T_n=a+(n-1)d\\T_{15}=-11+14*3\\=-11+42\\=31$

Hence, the 7th term from the end of the AP is 31.