Question

In an a p the sum of 11th term is 44 and the sum of next 11th term is 55 and find the c d and first term

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{First\:term=\frac{39}{11}}}}$

$\green{\tt{\therefore{Common\:difference=\frac{1}{11}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given: }} \\ \tt: \implies Sum \: of \: first \: 11th \: term(s_{11}) = 44 \\ \\ \tt: \implies Sum \: of \: next \: 11th \: term(s_{11}) \\\\ \red{\underline \bold{To\:Find: }}\\ \tt:\implies First\:term(a)=?\\\\ \tt:\implies Common\:difference(d)=?$

• According to given question :

$\tt \circ \: First \: term = a \\ \\ \tt \circ \: Common \: difference = d \\ \\ \tt \circ \: Number \: of \: terms = 11 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ \tt: \implies s_{11} = \frac{11}{2} (2a + (11 - 1)d) \\ \\ \tt: \implies 44 \times \frac{2}{11} = 2a + 10d \\ \\ \tt: \implies a + 5d = 4 - - - - - (1) \\ \\ \tt \circ \: First \: term = a_{12} = a + 11d \\ \\ \tt \circ \: Common \: difference = d \\ \\ \tt \circ \: Number \: of \: terms = 11 \\ \\ \bold{For \: sum \: of \: next \: 11th \: term} \\ \tt: \implies s_{11} = \frac{11}{2} (2 a_{12} + (11 - 1)d) \\ \\ \tt: \implies 55 \times \frac{2}{11} = 2(a + 11d) + 10d \\ \\ \tt: \implies 10 = 2a + 32d \\ \\ \tt: \implies a + 16d = 5 - - - - - (2) \\ \\ \text{Subtracting \: (1) \: from \: (2)} \\ \tt: \implies a + 16d - a - 5d = 5 - 4 \\ \\ \tt: \implies 11d = 1 \\ \\ \green{\tt: \implies d = \frac{1}{11} } \\ \\ \text{Putting \: value \: of \: d \: in \: (1)} \\ \tt: \implies a + 5 \times \frac{1}{11} = 4\\ \\ \tt: \implies 11a + 5 = 44 \\ \\ \tt: \implies 11a = 44 - 5 \\ \\ \green{ \tt: \implies \frac{39}{11} }$

Answer 2

Correct Question :-

In an a p the sum of 11th term is 44 and the sum of next 11th term is 55 and find the common difference and first term.

Solution :-

Let the first term be a. And the common difference be d. And given S11=44.

As we know,

$Sn = \frac{n}{2} (2a + (n - 1)d)$

So, the sum of 11th term,

$Sn = \frac{11}{2} (2a + (11 - 1)d)$

$S11 = \frac{11}{2} (2a + 10d)$

$44 \times \frac{2}{11} = (2a + 10d)$

$8 = (2a + 10d).............eq1$

And the sum of next 11 terms =55

So the total number of terms =11+11=22 terms

And the sum of 22 terms(S22) =S11+next S11=44+55=99.

Now,

$S22 = \frac{22}{2} (2a + (22 - 1)d)$

$S22 = 11(2a + 21d)$

$99 = 11(2a + 21d)$

$\frac{99}{11} = 2a + 21d$

$9 = 2a + 21d................eq2$

Now by subtracting eq1 from eq2, we get

$= > 21d - 10d = 1 \\ = > 11d = 1 \\ = > d = \frac{1}{11}$

And putting the value of d =1/11,in eq1 we get

$= > 2a + 10d = 8 \\ = > 2a + \frac{10}{11} = 8 \\ = > 2a = 8 - \frac{10}{11} \\ = > a = \frac{78}{22} \\ = > a = \frac{39}{11}$

Therefore the common difference is 1/11 and the first term is 39/11.