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Question
An arithmetic progression 5, 12, 19, …. has 50 terms. Find its last term. Hence find the sum of its last 15 terms.
Solution
5, 12, 19, …………50 terms
Common difference, d = 7
First term, a = 5
Last term, t50 = a + (50 – 1)d = 5 + (50 – 1) × 7 = 5 + 49 × 7 = 5 + 343 = 348
Sum of last 15 terms = S50 – S35
`=50/2[2xx5+[50-1]xx7]-35/2[2xx5+[35-1]xx7]`
`=25[10+343]-35/2[10+34xx7]`
`=25xx353-35/2xx248`
=8825-4340
=4485
The sum of last 15 terms = 4485.
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