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Question
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
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Solution
t18 = 52 and t39 = 148 S56 = ?
tn = a + (n-1)d
t18 = a +( 18 - 1) d
52 = a + 17 d ......(i)
t39 = a + (39 -1) d
∴ 148 = a + 38d ........(2)
Adding (1) and (2)
a + 17d = 52
a + 38d = 148
2a = 55d = 200 .......(3)
Sn = `n/2` [2a + (n - 1) d ]
S56 = `56/2` [2a + (56 -1) d]
= 28 [2a + 55d ]
= 28(200) [from eq (3)]
S56 = 5600
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