"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 :"

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Accepted Answer

"$${{{{10} \over 2}(2{a_1} + 9d)} \over {{p \over 2}(2{a_1} + (p - 1)d)}} = {{100} \over {{p^2}}}$$

$(2{a_1} + 9d)p = 10(2{a_1} + (p - 1)d)$

$9dp = 20{a_1} - 2p{a_1} + 10d(p - 1)$

$9p = (20 - 2p){{{a_1}} \over d} + 10(p - 1)$

${{{a_1}} \over d} = {{(10 - p)} \over {2(10 - p)}} = {1 \over 2}$

$\therefore$ $${{{a_{11}}} \over {{a_{10}}}} = {{{a_1} + 10d} \over {{a_1} + 9d}} = {{{1 \over 2} + 10} \over {{1 \over 2} + 9}} = {{21} \over {19}}$$"

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 :

Solution

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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 :

Solution

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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 :