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प्रश्न
Find the sum of the following APs.
`1/15, 1/12, 1/10`, ......, to 11 terms.
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उत्तर
`1/15, 1/12, 1/10,` ......, to 11 terms
For this A.P.,
a = `1/15`
n = 11
d = a2 - a1
= `1/12-1/15`
= `(5-4)/60`
= `1/60`
We know that
Sn = `n/2[2a + (n -1)d]`
S11 = `11/2[2(1/15)+(11-1)1/60]`
S11 = `11/2[2/5+10/60]`
S11 = `11.2[2/15+1/6]`
S11 = `11/2[(4+5)/30]`
S11 = `(11/2)(9/30)`
S11 = `33/20`
Thus, the required sum of first 11 terms is `33/20`.
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Q.17
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Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
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