A) 6, 53
B) 7, 54
C) 8, 86
D) 9, 90
Correct Answer: A
Solution :
(a): Here, first term a = 3 common difference d = 10 sum of n terms \[{{S}_{n}}=168\] \[\Rightarrow \frac{n}{2}[2\times 3+(n-1)10]=168\] \[\left[ \because {{S}_{n}}=\frac{n}{2}[2a+(n-1)]d \right]\] \[\Rightarrow \frac{n}{2}(6+10n-10)=168\] \[\Rightarrow \frac{n}{2}(10n-4)=168\] \[\Rightarrow n(5n-2)=168\] \[\Rightarrow 5{{n}^{2}}-2n-168=0\] \[\therefore \left( \because \text{By}\,\,\text{quadratic}\,\,\text{formula}\text{ }\times =\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right)\]\[=\frac{2\pm \sqrt{4+3360}}{10}=\frac{2\pm \sqrt{3364}}{10}=\frac{2\pm 58}{10}=\frac{60}{10}\] or \[\frac{-56}{10}\] Since, n cannot be negative, \[\therefore n=6\] So, \[{{a}_{6}}=a+(n-1)d\] \[\Rightarrow {{a}_{6}}=3+(6-1)\times 10=3+5\times 10=53\] Hence, \[n=6\] and \[{{a}_{6}}=53\]
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