Sum

Find 26^{th} term from last of an AP 7, 15, 23……., 767 consits 96 terms

Advertisement Remove all ads

#### Solution

**Method: I**

r^{th} term from end is given by

= T_{n} – (r – 1) d or

= (n – r + 1)^{th} term from beginning where n is total no. of terms.

m = 96, n = 26

∴ T_{26} from last = T_{(96-26+1)} from beginning

= T_71 from beginning

= a + 70d

= 7 + 70 (8) = 7 + 560 = 567

**Method: II**

d = 15 – 7 = 8

∴ from last, a = 767 and d = –8

`∴ T_26 = a + 25d = 767 + 25 (–8)`

= 767 – 200

= 567.

Concept: Arithmetic Progression

Is there an error in this question or solution?

Advertisement Remove all ads