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Find the sum of the first 25 terms of an A.P. whose second and third terms are 14 and 18 respectively. |
Answer» a2= 14 a + d = 14 ..... (i) a3= 18 a + 2d = 18 ...... (ii) Subtracting (i) from (ii), we get d = 4 Putting the value of d in (i), we get a = 14 – 4 = 10 \(S_{25} = \frac{25}{2}[2(a) + 24(d)]\) = 25 [58] = 1450 |
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