The 26^{text{th}},11^{text{th}},and the last | vedclass

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(C) Let the first term be $a$,common difference be $d$,and the number of terms be $n$.The $n^{	ext{th}}$ term of an $AP$ is given by $T_n = a + (n-1)

Vedclass · Accepted Answer

$-\frac{1}{5}$ and $27$

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(C) Let the first term be $a$,common difference be $d$,and the number of terms be $n$.The $n^{	ext{th}}$ term of an $AP$ is given by $T_n = a + (n-1)d$.Given that the $26^{	ext{th}}$ term is $0$:$a + 25d = 0$ .... $(i)$Given that the $11^{	ext{th}}$ term is $3$:$a + 10d = 3$ .... $(ii)$Subtracting equation $(ii)$ from equation $(i)$:$(a + 25d) - (a + 10d) = 0 - 3$$15d = -3$$d = -\frac{3}{15} = -\frac{1}{5}$Substituting $d = -\frac{1}{5}$ into equation $(i)$:$a + 25(-\frac{1}{5}) = 0$$a - 5 = 0$$a = 5$Given that the last term $l$ is $-\frac{1}{5}$:$l = a + (n-1)d$$-\frac{1}{5} = 5 + (n-1)(-\frac{1}{5})$Multiply the entire equation by $5$:$-1 = 25 - (n-1)$$-1 = 25 - n + 1$$-1 = 26 - n$$n = 27$Thus,the common difference is $-\frac{1}{5}$ and the number of terms is $27$.

The $26^{\text{th}}$,$11^{\text{th}}$,and the last term of an $AP$ are $0$,$3$,and $-\frac{1}{5}$,respectively. Find the common difference and the number of terms.

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