Question

4 th term of an AP is 11 and 8th term exceeds twice the 4th term by 5 find the sum of 10 terms​

Answer 1

Step-by-step explanation:

It is given that 4th term of AP is 11

$\sf a_4 = 11$

Now, by using formula a + (n - 1)d we get:

$\sf a_4 = a + (4 - 1)d$

$\sf a_4 = a + 3d$

$\sf a + 3d = 11 \:\:\:\Bigg\lgroup \bf{Equation\: (1)}\Bigg\rgroup$

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It is also given that 8th term exceeds twice the fourth term by 5.

$\sf a_{8} = 2a_{4} + 5$

Using nth term formula we get,

$\sf a + (8 - 1)d = 2a_{4} + 5$

$\sf a +7d = 2a_{4} + 5$

$\sf a + 7d = 2(a + 3d) + 5$

$\sf a + 7d = 2(11) + 5 \: \: \: \bigg \{putting \: the \: value \: of \: (a + 3d) = 11 \: from \: equation \: (1) \bigg \}$

$\sf a + 7d = 22 + 5$

$\sf a + 7d = 27 \:\:\:\Bigg\lgroup \bf{Equation\: (2)}\Bigg\rgroup$

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Now, substracting Equation (1) form Equation (2)

$\sf a + 7d - (a + 3d) = 27 - 11$

$\sf \cancel{a} + 7d - \cancel{a} - 3d = 27 - 11$

$\sf 4d= 16$

$\sf d = 4$

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Put d = 4 in equation (1) we get,

$\sf a + 3d = 11$

$\sf a + 3(4) = 11$

$\sf a + 12 = 11$

$\sf a = 11 - 12$

$\sf a = - 1$

Now, putting the value of a and d in the Genral form of AP we get:

-1 , -1 + 4, -1 + (2)(4), -1 + (3)(4)........

AP : -1, 3, 7, 11......

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Now, by using formula of sum of nth terms in AP we get:

$\sf S_n = \dfrac{n}{2} \bigg\{ 2 a + (n - 1) d\bigg\}$

$\sf S_{10} = \dfrac{10}{2} \bigg\{ 2 ( - 1) + (10 - 1)4\bigg\}$

$\sf S_{10}= 5 \bigg\{ - 2 + 9\times 4\bigg\}$

$\sf S_{10} = 5 \bigg\{-2 + 36\bigg\}$

$\sf S_{10} = 5 \big\{ 34\big\}$

$\sf S_{10} = 170$

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$\therefore\:\underline{\textsf{The sum of first 10 terms is \textbf{170}}}.$

Answer 2

Given :-

• 4 th term of an AP is 11
• 8th term exceeds twice the 4th term by 5

To Find :-

• sum.of 10 terms

Answer :-

• sum of first 10 terms of AP is 170

Solution :-

Formulae used :- $\sf{\bold{\green{\underline{\underline{a_n = a + ( n - 1 ) d }}}}}$

• 4th term of an AP is 11.

$\sf a_4 = a + ( 4 - 1 ) d$

$\sf 11 = a + ( 4 - 1 ) d$

$\sf 11 = a + 3d$

$\sf a + 3d = 11$ --- ( i )

• 8th term exceeds twice the 4th term by 5

$\sf a_8 = a + ( 8 - 1 ) d$

$\sf 11 \times 2 + 5 = a + ( 8 - 1 ) d$

$\sf 22 + 5 = a + 7d$

$\sf a + 7d = 27$------- ( ii )

• subtracting eq ( i ) from ( ii )

$\sf a + 7d - a - 3d = 27 - 11$

$\sf 4d = 16$

$\sf\bold d = 4$

• putting value of d in eq ( i )

$\sf a + 3d= 11$

$\sf a + 3 \times 4 = 11$

$\sf a = 11 - 12$

$\sf\bold a = - 1$

• finding sum of first 10 terms

formulae used : - $\sf{\bold{\green{\underline{\underline{\underline s_n = \dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}}}$

$\sf s_{10} = \dfrac{10}{2} [ 2\times - 1 + ( 10 - 1 ) 4 ]$

$\sf s_{10} = 5 [ -2 + 9\times 4 ]$

$\sf s_{10} = 5 [ -2 + 36 ]$

$\sf s_{10} = 5 \times 34$

$\sf s_{10} = 170$

$\sf{\underline{\boxed{\large{\blue{\mathsf{\dag s_{10}= 170 }}}}}}$