Question

The sum of first 7 terms of an A.P. is 182. If its 4th and 17th are in the ratio 1 :5, find the A.P.

2 , 10 , 18
5 , 10 , 15
16 , 30 , 58
-1 , -6 ,-11
​

Answer 1

GivEn:

• Sum of first 7 terms $\sf ( S_n )$ of an AP is 182.

• Ratio of 4th and 17th terms are 1:5.

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To find:

• Find AP.

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SoluTion:

$\underline{\bigstar\:\boldsymbol{As\:per\:given\: Question\::}}$

Ratio of 4th and 17th terms are 1:5.

$:\implies\sf \dfrac{(a + 3d)}{(a + 16d)} = \dfrac{1}{5}$

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$:\implies\sf 5(a + 3d) = a + 16d$

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$:\implies\sf 5a + 15d = a + 16d$

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$:\implies\sf - d = - 4a$

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$:\implies\bf \red{d = 4a}$⠀⠀⠀⠀⠀⠀⠀( eq. 1 )

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As we know that,

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

Therefore, the sum of 7 terms of AP is,

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$:\implies\sf 182 = \dfrac{7}{2} \bigg[ 2a + (7 - 1)d \bigg]$

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$:\implies\sf 182 = \dfrac{7}{2} \bigg[ 2a + 6d \bigg]$

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$:\implies\sf 182 \times 2 = 7 \bigg[ 2a + 6d \bigg]$

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$:\implies\sf \cancel{ \dfrac{364}{7}} = 2a + 6d$

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$:\implies\sf 52 = 2a + 6(4a)$⠀⠀⠀⠀⠀⠀⠀( from eq. 1 )

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$:\implies\sf 52 = 2a + 24a$

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$:\implies\sf 52 = 26a$

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$:\implies\sf a = \cancel{ \dfrac{52}{26}}$

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$:\implies{\underline{\boxed{\sf{\pink{a = 2}}}}}\;\bigstar$

━━━━━━━━━━━━━━━

★ Now, Put the value of a in eq. (1)

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$:\implies\bf d = 4 \times 2$

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$:\implies{\underline{\boxed{\sf{\purple{d = 8}}}}}\;\bigstar$

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Now, We can find the required AP -

• a = 2

• $\sf a_2$ = a + d = 2 + 8 = 10

• $\sf a_3$ = a + 2d = 2 + 2 × 8 = 18

• $\sf a_4$ = a + 3d = 2 + 3 × 8 = 26

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$\therefore$ Hence, 2, 10, 18, 26,..... is the required AP.

Thus, Option (I) is correct.

Answer 2

Solution :

$\underline{\bf{Given\::}}}}$

The sum of first 7 terms of an A.P. is 182. If it's 4th & 17th are in the ratio 1:5.

$\underline{\bf{Explanation\::}}}}$

Firstly, we know that formula of an A.P;

$\boxed{\bf{a_n=a+(n-1)d}}}$

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

A/q

$\mapsto\sf{a_4 :a_{17} = 1:5}\\\\\mapsto\sf{\dfrac{a+(4-1)d}{a+(17-1)d} =\dfrac{1}{5} }\\\\\\\mapsto\sf{\dfrac{a+3d}{a+16d} =\dfrac{1}{5} }\\\\\\\mapsto\sf{5(a+3d) = 1(a+16d) \:\:\underbrace{\bf{Cross-multiplication}}}\\\\\mapsto\sf{5a+15d = a+16d}\\\\\mapsto\sf{5a-a=16d-15d}\\\\\mapsto\sf{4a=d}\\\\\mapsto\bf{d=4a.................(1)}$

&

Using formula of the sum of an A.P;

$\boxed{\bf{S_n=\frac{n}{2} \bigg[2a+(n-1)d\bigg]}}}$

$\mapsto\sf{S_n=182}\\\\\mapsto\sf{182=\dfrac{7}{2} \bigg[2a+(7-1)(4a)\bigg]}\\\\\mapsto\sf{182=\dfrac{7}{2} \bigg[2a+28a-4a\bigg]}\\\\\mapsto\sf{182 = \dfrac{7}{2} \bigg[2a+24a\bigg]}\\\\\mapsto\sf{182 \times 2=7(26a)}\\\\\mapsto\sf{364=182a}\\\\\mapsto\sf{a=\cancel{364/182}}\\\\\mapsto\bf{a=2}$

∴Putting the value of a in equation (1),we get;

$\longrightarrow\sf{d=4\times 2}\\\\\longrightarrow\bf{d=8}$

Thus;

$\boxed{\bf{Arithmetic\:Progression\::}}}$

$\bullet\:\sf{a=\boxed{\bf{2}}}\\\\\bullet\sf{a+d=2+8 = \boxed{\bf{10}}}\\\\\bullet\sf{a+2d=2+2(8) =2+16= \boxed{\bf{18}}}$

Option (a)