Question
Find the 31st term of the series 25, 29,
33β¦β¦Solution
The given series is in the Arithmetic progression. having First term (a) = 25 Common difference (d) = 29 β 25 = 33 β 29 = 4 So, the nth term (an) = a + (n β 1) Γ d So,31st term = 25 + (31 β 1) Γ 4 Β = 25 + 30 Γ 4 =25 + 120 = 145
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