Question

Find the sum of the given series 105+112 +119+........+196

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Sum\:of\:14th\:term=2107}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies Series = 105 + 112 + 119 + .... + 196 \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies Sum \: of \: series = ?$

• According to given question :

$\tt \circ \: First \: term = 105 \\ \\ \tt \circ \: Common \: difference = 7 \\ \\ \tt \circ \: Last \: term = 196 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies a_{n} = a + (n - 1)d \\ \\ \tt: \implies 196 = 105 + (n - 1) \times 7 \\ \\ \tt: \implies 196 - 105 = (n - 1) \times 7 \\ \\ \tt: \implies \frac{91}{7} = n - 1 \\ \\ \tt: \implies \frac{91 + 7}{7} = n \\ \\ \green{\tt: \implies n = 14} \\ \\ \bold{For \: sum \: of \: 14th \: term : } \\ \tt: \implies s_{n} = \frac{n}{2}(a + l) \\ \\ \tt: \implies s_{14} = \frac{14}{2} (105 + 196) \\ \\ \tt: \implies s_{14} = 7 \times 301 \\ \\ \green{\tt: \implies s_{14} = 2107}$

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ sum \ of \ the \ series \ is \ 2107.}$

$\sf\orange{Given:}$

$\sf{\implies{First \ term(a)=105,}}$

$\sf{\implies{Common \ difference (d)=112-105=7,}}$

$\sf{\implies{Last \ term(t_{n}=196}}$

$\sf\pink{To \ find:}$

$\sf{Sum \ of \ the \ series.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{196=105+(n-1)\times7}}$

$\sf{\therefore{7(n-1)=196-105}}$

$\sf{\therefore{7(n-1)=91}}$

$\sf{\therefore{n-1=\frac{91}{7}}}$

$\sf{\therefore{n=13+1}}$

$\sf{\therefore{n=14}}$

__________________________________

$\boxed{\sf{S_{n}=\frac{n}{2}[t_{1}+t_{n}]}}$

$\sf{\therefore{S_{14}=\frac{14}{2}[105+196]}}$

$\sf{\therefore{S_{14}=7\times301}}$

$\sf{\therefore{S_{14}=2107}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ the \ series \ is \ 2107.}}}$