Question

find the sum of. 7+15+23+.......+225​

Answer 1

Question:

$Find\:the\:sum\:of. 7+15+23+.......+225$

To Find:

$\text{The sum of the numbers}$

We know:

This sum can be found quickly by taking the number n of terms being added i.e 8, multiplying by the sum of the first and last number in the no.i.e (7 + 225), and dividing by 2:

Concept:

Formula for finding the sum:

$\dfrac{n\big(a_{1} + a_{n}\big)}{2}$

Calculation:

$\dfrac{8\big(7 + 225}{2}\big)$

$\dfrac{8\big(332}{2}\big)$

$4\big(332\big)$

$1328$

Conclusion:

Hence the sum of the no.s is 1328...

Answer 2

Correct question

7+15+23+.....+223

Answer:

$\sf{The \ sum \ of \ the \ terms \ is \ 3220.}$

Given:

• $\sf{Here, \ t_{1}=7, \ t_{2}=15, \ t_{3}=23 \ and \ t_{n}=223}$

To find:

• Sum of terms.

Solution:

$\sf{t_{2}-t_{1}=15-7=8,}$

$\sf{t_{3}-t_{2}=23-15=8.}$

$\sf{\therefore{The \ given \ sequence \ is \ an \ AP}}$

$\sf{Here, \ a=7, \ d=8 \ and \ t_{n}=223}$

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

$\sf{\therefore{223=7+(n-1)\times8}}$

$\sf{\therefore{8(n-1)=223-7}}$

$\sf{\therefore{8(n-1)=216}}$

$\sf{\therefore{n-1=\frac{216}{8}}}$

$\sf{\therefore{n-1=27}}$

$\sf{\therefore{n=28}}$

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

$\sf{\therefore{S_{28}=\frac{28}{2}[7+223]}}$

$\sf{\therefore{S_{28}=14\times230}}$

$\sf{\therefore{S_{28}=3220}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ the \ terms \ is \ 3220.}}}$