If twice the 11^{th} term of an Arithmetic Pr | vedclass

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(C) Given the formula for the $n^{th}$ term of an Arithmetic Progression: $t_n = a + (n - 1)d$.For the $11^{th}$ term: $t_{11} = a + 10d$.For the $21^

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(C) Given the formula for the $n^{th}$ term of an Arithmetic Progression: $t_n = a + (n - 1)d$.For the $11^{th}$ term: $t_{11} = a + 10d$.For the $21^{st}$ term: $t_{21} = a + 20d$.According to the problem: $2t_{11} = 7t_{21}$.$2(a + 10d) = 7(a + 20d)$.$2a + 20d = 7a + 140d$.$5a + 120d = 0$.Dividing by $5$: $a + 24d = 0$.Now,the $25^{th}$ term is $t_{25} = a + (25 - 1)d = a + 24d$.Since $a + 24d = 0$,therefore $t_{25} = 0$.

If twice the $11^{th}$ term of an Arithmetic Progression is equal to seven times its $21^{st}$ term,then its $25^{th}$ term is .......

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