Question

Find the sum to first 20 terms of the series:; 11 + 17 + 23 + 29 ……
A) 1240
B) 1260
C) 1360
D) 1420

Answer 1

Answer :-

→ Option C → 1360.

Step-by-step explanation :-

Given :-

→ AP :- 11, 17, 23, 29 , ..... .

→ d = 17 - 11 = 6 .

→ n = 20.

To find :-

→ Sum of first 20 terms ( $S_{20}$ .

$\huge \pink{ \mid \underline{ \overline{ \sf Solution :- }} \mid}$

We have,

→ AP = 11, 17, 23, 29 , ..... .

Then, sum of nth term is given by:)

$\tt S_n = \frac{n}{2} [ 2a + ( n - 1) d ].$

Then,

$\tt S_{20} = \frac{20}{2} [ 2 \times 11 + ( 20 - 1) 6 ]. \\ \\ \sf = 10(22 + 19 \times 6). \\ \\ \sf = 10(22 + 114). \\ \\ \sf= 10 \times 136. \\ \\ \huge \boxed{ \orange{ \tt = 1360.}}$

Hence, it is solved.

Answer 2

$\bold{First \: we \: need \: to \ check }$$\bold{ whether \: the \: series \: is \: in \: AP \: or \: GP}$

$\mathsf{For \: common \:ratio=\dfrac{Next \: term }{previous \: term}}$

$\mathsf{r = \dfrac{17}{11}}$

$\mathsf{r = \dfrac{23}{17}}$

Since, common ratio differes therefore the given series is not in GP.

$\mathsf{For \: common \:difference = Next \: term - previous \: term}$

$\mathsf{17 -11 = 6}$

$\mathsf{23 -17 = 6}$

$\mathsf{29 -23= 6}$

Since, the common difference is same,

Therefore, the given series is in AP

$\bold{For \: an \: AP}$

$\mathsf{First \: term \: a = 11}$

$\mathsf{Common \: difference \: d = 6}$

$\bold{For \: sum \: of \: n \: terms \: of \: an \: AP}$

$\huge \boxed{S_n = \dfrac{n}{2} \times [2a + (n -1)d]}$

On placing values,

$\mathsf{S_{20} = \dfrac{20}{2} \times (22 + (20 -1)6}$

$\mathsf{S_{20 }= 10 \times (11 + 19 \times 6}$

$\mathsf{S_{20} = 10 \times (22+ 114}$

$\mathsf{S_{20} = 10 \times (136)}$

$\huge \boxed {\mathsf{S_{20} = 1360}}$

So, $\bold{OPTION \: C \: IS \: CORRECT}$