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in an AP if S5+S7=167 and S10=235 than find AP

Posted by Anant Thapa 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

According to the question, S₅ + S₇ = 167
{tex}\Rightarrow \frac{5}{2}[2 a+4 d]+\frac{7}{2}{/tex}[ 2a + 6d] = 167
{tex}\Rightarrow{/tex} 5(a + 2d) + 7(a + 3d) = 167
{tex}\Rightarrow{/tex} 12a + 31d = 167 ......(i)
and S₁₀= 235
{tex}\Rightarrow \quad \frac{10}{2}{/tex}[2a + 9d] = 235
{tex}\Rightarrow{/tex} 2a + 9d = {tex}\frac{235}{5}{/tex} = 47 ........(ii)
Multiplying eq. (ii) by 6 and then subtracting from (i), we have

d = {tex}\frac{-115}{-23}{/tex} = 5
From (ii), we get
2a + 9d = 47
{tex}\Rightarrow{/tex} 2a + 9(5) = 47
{tex}\Rightarrow{/tex} 2a = 47 - 45
{tex}\Rightarrow{/tex} 2a = 2 {tex}\Rightarrow{/tex} a = 1
Therefore, A.P is 1, (1 + 5), (1 + 5 + 5), (1 + 5+ 5+ 5)........
i.e, 1, 6, 11, 16......

0Thank You

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