Question

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

Answer 1

Answer:

common difference, d = 99 - 101= -2

Last term =47

Let the number of terms be n

Then,

47=101+(n-1) *(-2)

=101-2n+2

=103-2n

or, 2n=103-47

or, n = 56/2=28

No. of terms = 28.

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

$\textsf{In this case, we have to find the number of terms.}$

$\textsf{Here, a = 101, l = Tn = 47, d = 99 -101 = -2}$

$\mathsf{ \therefore \: 47 = 101(n - 1) \times ( - 2),where \: n = number \: of \: terms}$

$\mathsf {\implies \: 47 = 101 - 2n + 1}$

$\mathsf {\implies \: 2n = 103 - 47}$

$\mathsf {\implies \: 2n = 56}$

$\mathsf {\implies \: n = \frac{56}{2} }$

$\mathsf {\implies \: n = 28}$

$\mathsf {\therefore \: S_n = \frac{n}{2}(a + l) }$

$\mathsf{S_{28} = \frac{28}{2} (101 + 47)} \\ = 14 \times 18 \\ = \bf2072$