Question

The first term of two ap s are equal and the ratio of their common differences is 1 : 2 . if the 7th term of first ap and 21th ap are 23 and 125 respectively . find two ap s

Answer 1

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

$\green{\tt{\therefore{First\:A.P=5,8,11,14,..}}}$

$\green{\tt{\therefore{Second\:A.P=5,11,17,23,..}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Ratio \: of \: common \: difference = 1 : 2 \\ \\ \tt: \implies 7th \: term \: of \: first \:A.P = 23 \\ \\ \tt: \implies 21th \: term \: of \: second \:A.P = 125 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies First \: A.P =? \\ \\ \tt: \implies Second \:A.P =?$

• According to given question :

$\tt \circ \: a_{1} = a_{1_{o}} =a \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies d_{1} : d_{2} = 1 : 2 \\ \\ \tt: \implies \frac{d_{1}}{d_{2}} = \frac{1}{2} \\ \\ \tt: \implies d_{2} = 2 d_{1} - - - - - (1) \\ \\ \bold{For \: 7th \: term \: of \: first \: ap} \\ \tt: \implies a_{7} = 23 \\ \\ \tt: \implies a + 6 d_{1} = 23 - - - - - (2) \\ \\ \bold{For \:21st\: term : }\\ \\ \tt: \implies a_{21} = 125 \\ \\ \tt: \implies a + 20d_{2} = 125 \\ \\ \tt: \implies a + 20 \times 2 d_{1} = 125 \\ \\ \tt: \implies a + 40 d_{1} = 125 - - - - - (3)$

$\text{Subtracting \: (1) \: from \: (2)} \\ \tt: \implies 40 d_{1} - 6 d_{1} = 125 - 23 \\ \\ \tt: \implies 34 d_{1} = 102 \\ \\ \tt: \implies d_{1} = \frac{102}{34} \\ \\ \green{\tt: \implies d_{1} = 3} \\ \\ \text{Putting \: value \: of \:} d_{1} \text{ \: in \: (1)} \\ \tt: \implies d_{2} = 2 \times 3 \\ \\ \green{\tt: \implies d_{2} = 6} \\ \\ \text{Putting \: value \: of} \: d_{1} \: \text{ in \: (2)} \\ \tt: \implies a + 6 \times 3 = 23 \\ \\ \tt: \implies a = 23 - 18 \\ \\ \green{\tt: \implies a = 5} \\ \\ \green{\tt \therefore First \:A.P = 5,8,11,14,....} \\ \\ \green{\tt \therefore Second \: A.P= 5,11,17,23,....}$

Answer 2

$\huge\sf\pink{Answer}$

☞ First AP = 5,8,11,14...

☞ Second AP = 5,11,17,23...

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$\huge\sf\blue{Given}$

➳ First term of 2 Aps are same

➳ Ratio of their common difference = 1:2

➳ 7 th term of the first AP = 23

➳ 21 st term of the second AP = 125

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$\huge\sf\gray{To \:Find}$

➢ The two APs?

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$\huge\sf\purple{Steps}$

❍ Let the First Term of Both APs be a.

❍ Common Difference of First AP be d and, of Second AP be 2d.

☆ 7 th term of the first AP

$\dashrightarrow\sf T_n = a + (n - 1)d\\ \\ \dashrightarrow\sf T_7 = a + (7 - 1)d \\ \\\dashrightarrow\tt 23 = a + 6d \quad\dfrac{ \quad}{}eq.(1)$

☆ 21st term of Second AP,

$\dashrightarrow\sf T_n = a + (n - 1)d\\ \\ \dashrightarrow\sf T_{21} = a + (21 - 1)2d \\\\\dashrightarrow\sf 125 = a + 20 \times 2d\\\\\\\dashrightarrow\sf 125 = a + 40d\qquad\dfrac{\quad}{}eq.(2)$

☆ Subtracting eq(1) from eq(2)

$\leadsto\sf a + 40d = 125\\\\ \leadsto\sf a + 6d = 23\\ \dfrac{\qquad\qquad\qquad \qquad \qquad}{}\\\leadsto\sf (40d - 6d) = (125 - 23)\\ \\ \leadsto\sf 34d = 102\\ \\ \leadsto\sf d = \cancel\dfrac{102}{34}\\\\\leadsto\green{\sf d = 3}$

☆ Putting the Values in Equation 1

$\leadsto\sf 23 = a + 6d\\\\ \leadsto\sf 23 = a +6(3)\\\\ \leadsto\sf 23 = a + 18\\ \\ \leadsto\sf 23 - 18 = a\\ \\\leadsto\green{\sf a = 5}$

$\underline{\bullet\:\textsf{First Arithmetic Progresion :}} \\\\\longrightarrow \sf a, (a + d), (a +2d), (a +3d)..\\\\ \longrightarrow\sf 5, (5 + 3), (5 + 2(3)),(5 + 3(3))..\\\\ \longrightarrow\large\sf\orange{ 5, 8,11,14..}$

$\underline{\bullet\:\textsf{Second Arithmetic Progresion :}}\\\\\longmapsto \sf a, (a + 2d), (a +2(2d)), (a +3(2d))..\\\\\longmapsto\sf 5, (5 + 2(3)), (5 + 2(6)), (5 + 3(6))..\\\\ \longmapsto\large\sf\orange{ 5, 11,17, 23..}$

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