the first term of the first ap is 3 less than the first term of the second AP the 10th term of the first ap is 49 and the 5th term of the second ap is 27 find the two AP if their common difference is the same
Answers
Answer:
The first A.P is 4 , 9 , 14 ......so on ,
The second A.P is 7 , 12 , 17 , .......so on
Step-by-step explanation:
Given as :
The first term of the first AP is 3 less than the first term of the second AP
Let nth term of first AP = = a + (n - 1) d
So, first term of first AP = = a + (1 - 1 ) d
∴ = a + 0 × d
i.e = a
And
Let nth term of second AP = = A + (n - 1) D
So, first term of second AP = = A + (1 - 1 ) D
∴ = A + 0 × D
i.e = A
So, According to question
= - 3
Or, a = A - 3 ..........1
Again
10th term of the first AP is 49
i.e = a + (10 - 1) d
Or, a + 9 d = 49
i.e A - 3 + 9 d = 49
Or, A + 9 d = 52 ............2
And
The 5th term of the second AP is 27
i.e = A + (5 - 1) D
Or, 27 = A + 4 D
According to question, common difference of first and second AP are same
i.e d = D
So, 27 = A + 4 d ...........3
Solving eq 2 and eq 3
(A + 9 d) - ( A + 4 d) = 52 - 27
i.e (A - A) + (9 d - 4 d) = 25
or, 0 + 5 d = 25
∴ d =
i.e d = 5
So, Common difference for both A.P = d = 5
Put the value of d in eq 3
27 = A + 4 d
Or, A = 27 - 4 × 5
or, A = 27 - 20
∴ A = 7
So, first term for second A.P = A = 7
put the value of A in eq 1
a = A - 3
i.e a = 7 - 3
∴ a = 4
So, first term for first A.P = a = 4
For first A.P
first term = 4 + (1 - 1) 5 = 4 +0 = 4
second term = 4 + (2 - 1) 5 = 4 + 5 = 9
third term = 4 + (3 - 1) 5 = 14 ........so on
Again
For second A.P
first term = 7 + (1 - 1) 5 = 7 +0 = 7
second term = 7 + (2 - 1) 5 = 7 + 5 = 12
third term = 7 + (3 - 1) 5 = 17 ........so on
Hence, The first A.P is 4 , 9 , 14 ......so on , and second A.P is 7 , 12 , 17 , .......so on , Answer