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प्रश्न
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
योग
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उत्तर
Here, first term (A) = `(a - b)/(a + b)`
And common difference,
D = `(3a - 2b)/(a + b) - (a - b)/(a + b)`
= `(2a - b)/(a + b)`
∵ Sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
⇒ Sn = `n/2{2((a - b))/((a + b)) + (n - 1) ((2a - b))/((a + b))}`
= `n/2{(2a - 2b + 2an - 2a - bn + b)/(a + b)}`
= `n/2((2an - bn - b)/(a + b))`
∴ S11 = `11/2{(2a(11) - b(11) - b)/(a + b)}`
= `11/2((22a - 12b)/(a + b))`
= `(11(11a - 6b))/(a + b)`
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क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Arithematic Progressions - Exercise 5.3 [पृष्ठ ५३]
