Question

the sum of the first 7 A.P is 10 and that next 7 terms is 17. find the progression

Answer 1

Answer:

$Required \:A.P:\\1,\frac{8}{7},\frac{9}{7},...$

Step-by-step explanation:

Let a , d are first term and common difference of an A.P.

We know that,

$Sum \:of \:n\:terms (S_{n})\\=\frac{n}{2}[2a+(n-1)d]$

$i) Sum \:of \: first\:7\:terms=10$

$\implies \frac{7}{2}[2a+(6-1)d]=10$

$\implies \frac{7}{2}[2a+6d]=10$

$\implies 7(a+3d)=10$

$\implies 7a+21d = 10\:--(1)$

$Sum\:of \:next\:7\:terms = 17$

$\implies \frac{7}{2}[2(a+7d)+6d]=17$

$\implies \frac{7}{2}[2a+14d+6d]=17$

$\implies 7(a+10d)=17$

$\implies 7a+70d = 17\:--(2)$

/* Subtract equation (1) from equation (2), we get

$\implies 49d = 7$

$\implies d = \frac{7}{49}$

$\implies d = \frac{1}{7}$

/* Put d value in equation (1), we get

$7a+21\times \frac{1}{7}=10$

$\implies 7a + 3=10$

$\implies 7a = 7$

$\implies a = \frac{7}{7}=1$

$a = 1 , d =\frac{1}{7}$

Therefore,

Required A.P:

a = 1 ,

$a_{2}=a+d = 1+\frac{1}{7}=\frac{8}{7}$

$a_{3}=a_{2}+d=\frac{8}{7}+\frac{1}{7}=\frac{9}{7},...$

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