Question

The sum of the first 19 terms of an ap a1,a2.. If it is known that a4+a8+a12+a16=224

Answer 1

Step-by-step explanation:

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Answer 2

Concept:

A series of numbers is called a "arithmetic progression" (AP) when there is a consistent difference between any two subsequent numbers.

Given:

$a4+a8+a12+a16=224$

To find:

Sum of 19 terms of an AP

Solution:

The $n^{th}$ term and sum of $n^{th}$ is given as

 $\\ a_{n}=a+(n-1) d \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}) ,$

where d is common difference and a is the first term.

Since,

$a_{4}+a_{8}+a_{12}+a_{16}=224 \\\ a+3 d+a+7 d+a+11 d+a+15 d=224 \\ 4 a+36 d=224\\ a+9 d=56$

Now,    

Sum of the first 19 terms will be

$S_{19}=\frac{19}{2}[2 a+(19-1) d] \\&=19[a+9 d]$

Sustituting the value of $a+9d$

$=19[56] \\&=1064\end{aligned}$        

The sums of first 19 terms is 1064

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