The question is given from NCERT Book of class 10th Ch. no. 5 Ex. 5.2 Q. 8. In the following question you have to find the 29th term of an A.P which has 50 terms in which 3rd term is 12 and last term is 106. Give the answer briefly.
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An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
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Solution:
3rd term, a3 = 12
50th term, a50 = 106
We know,
an = a + (n−1)d
a3 = a + (3−1)d
12 = a + 2d ……………………………. (i)
Like above,
a50 = a+(50−1)d
106 = a+49d …………………………. (ii)
By subtracting (i) from (ii),
94 = 47d
d = 2 = common difference
From equation (i),
12 = a + 2(2)
a = 12 − 4 = 8
a29 = a + (29−1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
The 29th term is 64.