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Question
The 16th term of an A.P. is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.
Solution
Given that 16th term of an A.P. is five times its third term.
We know that
Thus,
`t_16=a+(16-1)d`
`t_3=a+(3-1)d`
Since `t_16= 5t_3 `, we have,
a+(16-1)d=5[a+(3-1)d]
a+15d=5[a+2d]
a+15d=5a+10d
5d=4a
4a-5d=0.......(1)
Also given that ` t_10= 41`
`t_10=a+(10-1)d`
41=a+9d
a+9d=41......(2)
Multiplying equation (2) by 4, we have,
4a + 36d = 164...(3)
Subtracting equation (1) from equation (3), we have,
[36-(-5)]d=164
41d=164
d=164/41
d=4
Substituting d = 4 in equation (1) 4a - 5d = 0,we have,
4a - 5 x 4=0
4a- 20= 0
4a=20
a=5
We need to find `S_15`
We know that
`S_n=n/2[2a+(n-1)d]`
`S_15=15/2[2xx5+(15-1)xx4][a=5,n=15,d=4]`
`S_15=15/2[10+14xx4]`
`S_15=15/2xx66`
`S_15=495`
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