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If the ratio of the sum of n terms of two AP's be (7n + 1):(4n + 27), then the ratio of their 11th terms will be
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a
2:3
b
3:4
c
4:3
d
5:6
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Correct option is C
Let Sn, and S' be the sums of n terms of two AP's and T11 and T11, be the respective 11th term, thenSnSn′=n2[2a+(n−1)d]n22a′+(n−1)d′=7n+14n+27 (given)⇒a+(n−1)2da′+(n−1)2d′=7n+14n+27 Now, put n=21, we get a+10da′+10d′=T11T11′=148111=43Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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