Question

if s4=1/27 and s7=1/729 find s11​

Answer 1

Given,

$\boxed{ \bf{}S_{\tiny4} = \frac{1}{27} \: } \: \: \: \: \: \: \: \: \: \sf \& \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{}S_{\tiny7} = \frac{1}{729} }$

• To Find : $\rm S_{\tiny{11}}$

We Know,

$\boxed{ \: \bullet \: \bf \: \: { S \gray{ =} \frac{1}{2} \{2a + (n - 1)d \}} \: \: }$

Thus,

$\hookrightarrow \: \sf{}S_{\tiny4} = \frac{1}{2} \{ 2a + (n - 1)d\} \\ \\ \tt \to \frac{1}{27} = \frac{1}{2} \{ 2a + (n - 1)d\} \\ \\ \to \tt \frac{ \frac{1}{27} }{ \frac{1}{2} } = \{ 2a + (n - 1)d\} \\ \\ \to \tt \frac{2}{27} = \{ 2a + (n - 1)d\} \\ \\ \to \tt \frac{2}{27} = \{ 2a + (4 - 1)d\} \\ \\ \to \tt \frac{2}{27} = \{ 2a + (3)d\} \\ \\ \to \tt \frac{2}{27} = 2a + 3d \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: \: \: - - (1)$

And,

$\hookrightarrow \: \sf{}S_{\tiny7} = \frac{1}{2} \{ 2a + (n - 1)d\} \\ \\ \tt \to \frac{1}{729} = \frac{1}{2} \{ 2a + (n - 1)d\} \\ \\ \to \tt \frac{ \frac{1}{729} }{ \frac{1}{2} } = \{ 2a + (n - 1)d\} \\ \\ \to \tt \frac{2}{729} = \{ 2a + (n - 1)d\} \\ \\ \to \tt \frac{2}{729} = \{ 2a + (7- 1)d\} \\ \\ \to \tt \frac{2}{729} = \{ 2a + (6)d\} \\ \\ \to \tt \frac{2}{729} = 2a + 6d \: \: \: \: \: \: \: \: \: \: \: \rm \: \: \: - - (2)$

Subtracting Equation (ii) from (i), we get,

$\sf{}2a + 6d - (2a + 3d) = \frac{2}{729} - \frac{2}{27} \\ \\ \sf \to2a + 6d - 2a - 3d = \frac{2 - 54}{729} \\ \\ \sf \to 3d = \frac{ - 52}{729} \\ \\ \sf \to{}d = \frac{ - 52}{729 \times 3} \: \to \: \green{ \frac{ - 52}{2187} }$

Using Equation (i),

$\sf \: a = \frac{ \frac{2}{27} - 3d}{2} \\ \\ \sf a = \frac{ \frac{2}{27} - 3( \frac{ - 52}{2187} ) }{2} \\ \\ \sf a = \frac{ \frac{2}{27} - \frac{ - 52}{729} }{2} \\ \\ \sf{}a = \frac{ \frac{54 - ( - 52)}{729} }{2} = \frac{ \frac{54 + 52}{729} }{2} = \frac{ \frac{106}{729} }{2} \\ \\ \sf a = \frac{106 \times 2}{729} = \green{\frac{212}{729}}$

Therefore,

$\huge \boxed{ \bf{}a = \pink{ \frak{ \frac{212}{729} }} \: \: \: }\\ \huge \boxed{\bf{d = \pink{ \frak{ \frac{ - 52}{2187}}} } }$

Hence,

$\sf{}S_{ \tiny{11}} = \frac{1}{2} \{ 2a + (n - 1)d\} \\ \\ \to \bf \frac{1}{2} \bigg\{ 2\bigg( \frac{212}{729} \bigg)+ (11 - 2) \bigg( \frac{ - 52}{2187} \bigg) \bigg\} \\ \\ \to \bf \frac{1}{2} \bigg\{ 2\bigg( \frac{212}{729} \bigg)+ (7) \bigg( \frac{ - 52}{2187} \bigg) \bigg\} \\ \\ \to \bf \frac{1}{2} \bigg\{ 2\bigg( \frac{212}{729} \bigg)+ 7 \bigg( \frac{ - 52}{2187} \bigg) \bigg\} \\ \\ \bf \to \frac{1}{\cancel2} \times \cancel2 \bigg( \frac{212}{729} \bigg) + \frac{1}{ 2} \times 7 \bigg( \frac{ - 52}{2187} \bigg) \\ \\ \to \bf \frac{212}{729} + \frac{7 \times ( - 26)}{2187} \\ \\ \to \bf \frac{212 + ( - 182)}{2187} = \frac{212 - 182}{2187} \\ \\ \to \sf \frac{30}{2187} \sim \red{\frac{10}{729} }$

Thus,

$\huge \sf \bigstar \: \: S_{ \small{11}} = \red{ \tt\frac{10}{729} }$