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

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(B) The given series is an arithmetic progression: $101, 99, 97, \dots, 47$.Here,the first term $a = 101$,the common difference $d = 99 - 101 = -2$,an

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$28$

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(B) The given series is an arithmetic progression: $101, 99, 97, \dots, 47$.Here,the first term $a = 101$,the common difference $d = 99 - 101 = -2$,and the last term $l = 47$.We use the formula for the $n^{th}$ term of an arithmetic progression: $T_n = a + (n - 1)d$.Substituting the values,we get: $47 = 101 + (n - 1)(-2)$.Subtracting $101$ from both sides: $47 - 101 = (n - 1)(-2)$.$-54 = (n - 1)(-2)$.Dividing by $-2$: $27 = n - 1$.Therefore,$n = 28$.

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

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