Two A.Ps have the same common difference. The first term of one of these A.Ps is 3, and that of the other is 8. What is the difference between their second and the fourth terms?
[a] 5,10
[b] 5,5
[c] 5,7
[d] 5,14
Answer
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Hint: Assume that the common difference of the A.Ps is d. Using ${{a}_{n}}=a+\left( n-1 \right)d$, find the second terms of both the A.Ps and hence find the difference of the second terms of the A.Ps. Again using ${{a}_{n}}=a+\left( n-1 \right)d$, find the fourth terms of both the A.Ps and hence find the difference of the fourth terms of the A.Ps. Alternatively, use the fact that the sequence formed by subtracting respective terms of two A.Ps is also an A.P with common difference as the difference between the common differences of the A.Ps and the first term as the difference of the first terms of the two A.Ps.
Complete step-by-step answer:
Let d be the common difference of the two A.Ps.
We know that ${{a}_{n}}=a+\left( n-1 \right)d$
Put n = 2 and a = 3 to get the second term of the first A.P
We have ${{a}_{2}}=3+\left( 2-1 \right)d=3+d$
Again put n = 2 and a = 8 to get the second term of the second A.P
We have ${{A}_{2}}=8+\left( 2-1 \right)d=8+d$
Hence we have ${{A}_{2}}-{{a}_{2}}=8+d-\left( 3+d \right)=8+d-3-d=5$
Put n = 4 and a = 3 to get the fourth term of the first A.P
We have ${{a}_{4}}=3+\left( 4-1 \right)d=3+3d$
Put n = 4 and a = 8 to get the fourth term of the second A.P
We have ${{A}_{4}}=8+\left( 4-1 \right)d=8+3d$
Hence we have ${{A}_{4}}-{{a}_{4}}=8+3d-3-3d=5$
Hence the difference between the second terms and the fourth terms is 5,5 respectively.
Hence option [b] is correct.
Note: Alternatively, we have
First term of the A.P formed by subtracting respective terms of the A.Ps = 8-3 = 5
Also, the common difference of the A.P = d-d = 0.
Hence the difference of the second terms $5+1\times 0=5$ and the difference of the fourth terms $=5+3\times 0=5$, which is the same as obtained above.
Complete step-by-step answer:
Let d be the common difference of the two A.Ps.
We know that ${{a}_{n}}=a+\left( n-1 \right)d$
Put n = 2 and a = 3 to get the second term of the first A.P
We have ${{a}_{2}}=3+\left( 2-1 \right)d=3+d$
Again put n = 2 and a = 8 to get the second term of the second A.P
We have ${{A}_{2}}=8+\left( 2-1 \right)d=8+d$
Hence we have ${{A}_{2}}-{{a}_{2}}=8+d-\left( 3+d \right)=8+d-3-d=5$
Put n = 4 and a = 3 to get the fourth term of the first A.P
We have ${{a}_{4}}=3+\left( 4-1 \right)d=3+3d$
Put n = 4 and a = 8 to get the fourth term of the second A.P
We have ${{A}_{4}}=8+\left( 4-1 \right)d=8+3d$
Hence we have ${{A}_{4}}-{{a}_{4}}=8+3d-3-3d=5$
Hence the difference between the second terms and the fourth terms is 5,5 respectively.
Hence option [b] is correct.
Note: Alternatively, we have
First term of the A.P formed by subtracting respective terms of the A.Ps = 8-3 = 5
Also, the common difference of the A.P = d-d = 0.
Hence the difference of the second terms $5+1\times 0=5$ and the difference of the fourth terms $=5+3\times 0=5$, which is the same as obtained above.
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