The number of terms in the series 101 + 99 + | vedclass

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Question

(B) The given series is $101, 99, 97, \dots, 47$.This is an arithmetic progression $(AP)$ where the first term $a = 101$ and the common difference $d

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(B) The given series is $101, 99, 97, \dots, 47$.This is an arithmetic progression $(AP)$ where the first term $a = 101$ and the common difference $d = 99 - 101 = -2$.The last term $l$ (or $T_n$) is $47$.The formula for the $n^{th}$ term of an $AP$ is $T_n = a + (n - 1)d$.Substituting the values,we get $47 = 101 + (n - 1)(-2)$.Subtracting $101$ from both sides,we get $47 - 101 = (n - 1)(-2)$,which simplifies to $-54 = (n - 1)(-2)$.Dividing both sides by $-2$,we get $27 = n - 1$.Therefore,$n = 27 + 1 = 28$.Thus,the number of terms in the series is $28$.

The number of terms in the series $101 + 99 + 97 + ..... + 47$ is

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