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Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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Given that,

Second term, a2 = 14

Third term, a3 = 18

Common difference, d = a3−a2 

= 18−14 = 4

a2 = a+d

14 = a+4

a = 10 = First term

Sum of n terms;

Sn = n/2 [2a + (n – 1)d]

S51 = 51/2 [2×10 (51-1) 4]

= 51/2 [2+(20)×4]

= 51 × 220/2

= 51 × 110

= 5610

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