Question

If sum of first 6 terms of an A.P is 36 and that of the first 16 terms is 256.Find the sum of first 11 terms.​

Answer 1

$\Large{\underline{\underline{\mathfrak{\red{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\orange{\bf{Given}}}}}$

• Sum of first 6 terms of an A.p is 36

• First 16 terms is 256

$\Large{\underline{\mathfrak{\orange{\bf{Find}}}}}$

• Sum of first 11 terms.

$\Large{\underline{\underline{\mathfrak{\red{\bf{Explanation}}}}}}$

★ Sum of n terms = n/2(2a+(n-1)×d)

According to question!

Case (I).

Sn = n/2(2a+(n-1)×d)

Putting values,

➡ S6= 6/2(2a+(6-1)×d)

➡ 36/3 = (2a+6d-d)

➡12 = (2a+5d)

➡ 2a+5d= 12 ______________(1)

Case (II).

Sn = n/2 (2a+(n-1)×d

Putting value,

➡S16 = 16/2(2a+(16-1)×d)

➡ 256/8= 2a + 16d -d

➡ 32 = 2a+15d

➡2a+15d =32_____________(2)

Solve equation (1) and (2)

➡ 10d = 20

➡d= 20/10

➡d= 2. [ Common Difference ]

Putting value of d in eq (1)

• 2a+5d =12

➡2a +5×2=12

➡2a +10=12

➡ 2a=12-10

➡2a = 2

➡a= 2/2

➡ a= 1. [ First terms ]

Now

★Sum of 11 terms = n/2(2a+(n-1)×d)

➡ S11 = 11/2(2×1+(11-1)×2)

➡ S11= 5.5(2+ 10×2)

➡ S11= 5.5× 22

➡S11 = 121

$\Large{\underline{\underline{\mathfrak{\red{\bf{Hence}}}}}}$

• Sum of First 11 term is 121.

____________________________

Answer 2

${ \huge{ \underline{ \underline{ \sf{ \purple{AnswEr :}}}}}}$

• Sum of 11 terms is 121

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${ \huge{ \underline{ \underline{ \sf{ \orange{GivEn : }}}}}}$

• Sum of first 6 terms = 36
• Sum of first 16 terms = 256

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${ \huge{ \underline{ \underline{ \sf{ \blue{To \: find :}}}}}}$

• sum of first 11 terms = ??

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${ \huge{ \underline{ \underline{ \sf{ \pink{SoluTion : }}}}}}$

$\sf{\red{\boxed{\bold{S = \dfrac{n}{2}[2a + ( n - 1 ) d]}}}}$

☆$:\sf\implies 36 = \dfrac{6}{2}[ 2a + ( 6 - 1 )d ]$

$:\sf\implies 36 = \dfrac{6}{2}[ 2a + 5d ]$

$:\sf\implies 36 \times\dfrac{2}{6} = [ 2a + 5d ]$

$:\sf\implies 6\times 2 = 2a + 5d$

$\sf{\bold{\green{\underline{\underline{2a + 5d = 12 }}}}}$----- ( i )

☆$:\sf\implies 256 = \dfrac{16}{2}[ 2a + ( 16 - 1 )d ]$

$:\sf\implies 256 = \dfrac{16}{2}[ 2a + 15d ]$

$:\sf\implies 256 \times\dfrac{2}{16} = [ 2a + 15d ]$

$:\sf\implies 16\times 2 = 2a + 15d$

$\sf{\bold{\green{\underline{\underline{2a + 15d = 32 }}}}}$----- ( ii )

• subtract eq i from ii

$\sf\longrightarrow 2a + 15d - 2a - 5d = 32 - 12$

$\sf\longrightarrow 10d = 20$

$\sf\longrightarrow d = \dfrac{20}{10}$

$\sf\longrightarrow d = 2$

$\sf{\orange{\boxed{\bold{common\:Difference = 2 }}}}$

• putting value of d in eq i

$\sf\longrightarrow 2a + 5d = 12$

$\sf\longrightarrow 2a + 5\times 2 = 12$

$\sf\longrightarrow 2a + 10 = 12$

$\sf\longrightarrow 2a = 12 - 10$

$\sf\longrightarrow 2a = 2$

$\sf\longrightarrow a = 1$

$\sf{\orange{\boxed{\bold{a = 1 }}}}$

• finding sum of 11 terms

$:\sf\leadsto S_{11} = \dfrac{11}{2}[ 2\times 1 + ( 11 - 1 )2 ]$

$:\sf\leadsto S_{11} = \dfrac{11}{2}[ 2 + ( 10\times 2) ]$

$:\sf\leadsto S_{11} = \dfrac{11}{2} [ 2 + 20 ]$

$:\sf\leadsto S_{11} = \dfrac{11}{2}\times 22$

$:\sf\leadsto S_{11} = 11 \times 11$

$:\sf\leadsto S_{11} = 121$

$\sf{\red{\boxed{\bold{S_{11} = 121}}}}$

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