Q. 10 A4.2( 10 Votes )

Find the 10th term form the end of the AP 4, 9, 14, .. , 254.

The above series of numbers forms an arithmetic progression with

first term(a) = 4 and,

common difference(d) = (n + 1)th term – nth term = 9 – 4 = 5

last term or nth term(an) = 254

Let the total no. of terms in above A.P be n.

an = a + (n – 1) × d

254 = 4 + (n – 1) × 5

250 = 5n – 5

5n = 255

n = 51

10th term from the end of AP = 51 – 10 + 1 = 42th term from the beginning

42th term = a42 = a + (42 – 1)d

= 4 + 41 × 5

= 209

Hence, 10th term from the end of AP is 209.

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