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Find the sum of first 40 terms of an AP whose 2nd term is 2 and 9th term is 37
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S40 = ? , a2 = 2 ,a9 = 37
a2 = a + d = 2 ---- 1
a9 = a + 8d = 37---- 2
solving 1 and 2
a + d = 2
a + 8d = 37
- - -
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- 7d= -35
d = 35/7
d = 5
Substitute the value of d = 5 in equation 1
a + d = 2
a + 5 = 2
a = 2 - 5
a = -3
Sn = n/2{2a +( n -1)d}
S40 = 40/2 {2(-3) + (40 - 1)5}
= 20{-6 + 195}
= 20( 189)
S40 = 3780
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