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Question
In an A.P. t10 = 57 and t15 = 87, then find t21
Solution
For an A.P., let a be the first term and d be the common difference.
t10 = 57 and t15 = 87 ......[Given]
Since tn = a + (n – 1)d
t10 = a + (10 – 1)d
∴ 57 = a + 9d
i.e., a + 9d = 57 ......(i)
Also, t15 = 87
∴ a + (15 – 1)d = 87
∴ a + 14d = 87 ......(ii)
Subtracting equation (i) from (ii), we get
a + 14d = 87
a + 9d = 57
– – –
5d = 30
∴ d = `30/5` = 6
Substituting d = 6 in equation (i), we get
a + 9(6) = 57
∴ a + 54 = 57
∴ a = 57 – 54 = 3
t21 = a + (21 – 1)d
= a + 20d
= 3 + 20(6)
= 3 + 120
∴ t21 = 123
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