JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 11)
Let a1, a2, a3, ..... be an A.P. If $${{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}$$, p $$\ne$$ 10, then $${{{a_{11}}} \over {{a_{10}}}}$$ is equal to :
$${{19} \over {21}}$$
$${{100} \over {121}}$$
$${{21} \over {19}}$$
$${{121} \over {100}}$$
Explanation
$${{{{10} \over 2}(2{a_1} + 9d)} \over {{p \over 2}(2{a_1} + (p - 1)d)}} = {{100} \over {{p^2}}}$$
$$(2{a_1} + 9d)p = 10(2{a_1} + (p - 1)d)$$
$$9dp = 20{a_1} - 2p{a_1} + 10d(p - 1)$$
$$9p = (20 - 2p){{{a_1}} \over d} + 10(p - 1)$$
$${{{a_1}} \over d} = {{(10 - p)} \over {2(10 - p)}} = {1 \over 2}$$
$$\therefore$$ $${{{a_{11}}} \over {{a_{10}}}} = {{{a_1} + 10d} \over {{a_1} + 9d}} = {{{1 \over 2} + 10} \over {{1 \over 2} + 9}} = {{21} \over {19}}$$
$$(2{a_1} + 9d)p = 10(2{a_1} + (p - 1)d)$$
$$9dp = 20{a_1} - 2p{a_1} + 10d(p - 1)$$
$$9p = (20 - 2p){{{a_1}} \over d} + 10(p - 1)$$
$${{{a_1}} \over d} = {{(10 - p)} \over {2(10 - p)}} = {1 \over 2}$$
$$\therefore$$ $${{{a_{11}}} \over {{a_{10}}}} = {{{a_1} + 10d} \over {{a_1} + 9d}} = {{{1 \over 2} + 10} \over {{1 \over 2} + 9}} = {{21} \over {19}}$$
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