1.

Which term of the arithmetic progression 8, 14, 20, 26,..... will be 72 more than its 41th term?

Answer»

a = 8, d = 6,

Last term = an

an = a + (n – 1) d

= 8 + (n – 1) 6

= 2 + 6n (i)

Let, 41st term, a41 = 8 + (41 – 1) 6

= 8 + 40 x 6 = 248

Now the term is 72 more than its 41st term

an = 72 + a41

= 72 + 248 = 320

Putting this value in (i), we get

320 = 2 + 6n

6n = 318

n = 53

Hence, 53rd term of the given A.P. is 72 more than its 41st term



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