Question

Find the sum of first ten terma of an arithmetic progression whos fourth term is 13 and eighth term is 29​

Answer 1

Given : -

4th term = 13

8th term = 29

Required to find : -

• Sum of first 10 terms ?

Formula used : -

To find the sum of first nth terms we use the formula ;

$\Large{\dag{\boxed{\tt{\bf{ {S}_{nth} = \dfrac{ n}{ 2 } [ 2a + ( n - 1 )d ] }}}}}$

Here,

a = first term

d = common difference

n = term number

Solution : -

4th term = 13

8th term = 29

we need to find the sum of first 10 terms

The 4th term can be represented as ;

4th term = a + 3d

Similarly,

The 8th term can be represented as ;

8th term = a + 7d

This implies ;

a + 3d = 13 $\tt{\red{\bf{ Equation - 1 }}}$

Consider this as equation - 1

a + 7d = 29 $\tt{\red{\bf{ Equation - 2 }}}$

Consider this as equation - 2

Now,

Let's solve these 2 equations simultaneously ;

So,

we shall use the Elimination method to find the values of a & d .

Subtract equation 1 from equation 2

$\tt a + 7d = 29 \\ \tt a + 3d = 13 \\ \underline{ ( - )( - ) \: \: ( - ) \: \: } \\ \tt \underline{ \: \: \: \: \: + 4d = 16 \: } \\ \\ \implies \tt 4d = 16 \\ \\ \implies \tt d = \dfrac{16}{4} \\ \\ \implies \tt d = 4$

Substitute the value of d in Equation 1

=> a + 3d = 13

=> a + 3(4) = 13

=> a + 12 = 13

=> a = 13 - 12

=> a = 1

Hence,

• First term ( a ) = 1

• Common difference ( d ) = 4

Now,

Let's find the sum of first 10 terms ;

Using the formula ,

$\Large{\dag{\boxed{\tt{\bf{ {S}_{nth} = \dfrac{ n}{ 2 } [ 2a + ( n - 1 )d ] }}}}}$

$\: \longrightarrow \tt {S}_{nth} = {S}_{10} \\ \\ \longrightarrow \tt {S }_{10} = \dfrac{10}{2} [ 2(1) + ( 10 - 1 )4 ] \\ \\ \longrightarrow \tt {S}_{10} = \dfrac{10}{2} [ 2 + ( 9 ) 4 ] \\ \\ \longrightarrow \tt {S}_{10} = 5 [ 2 + ( 9 ) 4 ] \\ \\ \longrightarrow \tt {S}_{10} = 5 [ 2 + 36 ] \\ \\ \longrightarrow \tt {S}_{10} = 5 [ 38 ] \\ \\ \longrightarrow \tt {S}_{10} = 5 \times 38 \\ \\ \longrightarrow \tt {S}_{10} = 190$

Therefore ,

Sum of first 10 terms = 190

Answer 2

GivEn:-

• $\sf 4^{th}$ term of AP $\sf ( a_{4} )$ = 13

• $\sf 8^{th}$ term of AP $\sf ( a_{8} )$ = 29

To find:-

• Sum of first 10 terms of An AP.

SoluTion:-

We have,

✇ $\sf 4^{th}$ term of AP $\sf ( a_{4} )$ = 13

$:\implies$ a + 3d = 13 ----(1)

✇ $\sf 8^{th}$ term of AP $\sf ( a_{8} )$ = 29

$:\implies$ a + 7d = 29 ----(2)

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✠ Using Substitution method -

★ From eq(1) -

$:\implies\sf \underline{a + 3d = 13}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a = 13 - 3d\;----(3)$

⠀⠀⠀⠀⠀⠀⠀

★ Put eq(3) in eq(2) -

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf (13 - 3d) + 7d = 29$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 13 - 3d + 7d = 29$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 13 + 4d = 29$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 4d = 29 - 13$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 4d = 16$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf d = \cancel{ \dfrac{16}{4}}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf{\underline{\boxed{\bf{\pink{d = 4}}}}}$ ★

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★ Now, Put value of d in eq(3) -

$:\implies\sf a = 13 - 3(4)$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a = 13 - 12$

⠀⠀⠀⠀⠀⠀⠀

$:\implies{\underline{\boxed{\bf{\blue{a = 1}}}}}$ ★

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★ Now, We have to find,

☆ Sum of 10 terms of an AP -

As we know that,

$\star\;\normalsize{\underline{\boxed{\bf{\purple{ S_n = \dfrac{n}{2} \bigg[ 2a + (n - 1)d \bigg]}}}}}$

⠀⠀⠀⠀⠀⠀⠀

Here, we have to find $\sf S_n$ -

$\;\;\small\sf\dag\; \underline{Put\;the\;givEn\;values\;in\;above\;formula\;-}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf S_{10} = \dfrac{10}{2} \bigg[ 2 \times 1 + (10 - 1)4 \bigg]$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf S_{10} = 5( 2+ 9 \times 4)$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf S_{10} = 5( 2 + 36)$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf S_{10} = 5(38)$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf S_{10} = 5(38)$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf{\underline{\boxed{\bf{\orange{S_{10} = 190}}}}}$ ★

⠀⠀⠀⠀⠀⠀⠀

$\therefore\;\sf \underline{ Sum\;of\;10\; terms \;of\; an \;AP \;is\; \bf{190.}}$

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★ Formula Used -

$\begin{lgathered}\boxed{\begin{minipage}{10 em} \sf \displaystyle \bullet a_n=a + (n-1)d \\\\\\ \bullet S_n= \dfrac{n}{2} \left(a + a_n\right) \end{minipage}}\end{lgathered}$

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