Question

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

Answer 1

Answer: for first A.P a₁=5 and d₁=3

for second A.P a₂=5 and d₂=6

Step-by-step explanation: a₁ = a₂=a   and d₁/d₂ = 1/2=1x/2x

therefore d₁=x   d₂=2x  

a₁+(7-1)d₁ = a+6x=23_____________________________eq 1

a₂+(21-1)d₂ = a+(20)2x = a+40x=125_________________eq 2

now on solving eq 1 and 2

a=5 and x=3

therefore for first A.P a₁=5 and d₁=3

for second A.P a₂=5 and d₂=6

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Answer 2

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

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

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

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

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

• According to given question :

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

$\text{Subtracting \: (1) \: from \: (2)}$

$\tt: \implies 40 d_{1} - 6 d_{1} = 125 - 23 \\ \\ \tt: \implies 34 d_{1} = 102 \\ \\ \tt: \implies d_{1} = \frac{102}{34} \\ \\ \green{\tt: \implies d_{1} = 3} \\ \\ \text{Putting \: value \: of \:} d_{1} \text{ \: in \: (1)} \\ \tt: \implies d_{2} = 2 \times 3 \\ \\ \green{\tt: \implies d_{2} = 6} \\ \\ \text{Putting \: value \: of} \: d_{1} \: \text{ in \: (2)} \\ \tt: \implies a + 6 \times 3 = 23 \\ \\ \tt: \implies a = 23 - 18 \\ \\ \green{\tt: \implies a = 5}$

Hence the answer is

• ∴First A.P =5,8,11,14,.
• ∴Second A.P =5,11,17,23,.