One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.
In this question we have been given that the sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161
Now we have to find the 28th term of this A.P.
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 30
Sum of first 7 terms of an A.P., S7 = 63.
And sum of next 7 terms is 161.
So, the sum of first 14 terms, S14 = Sum of first 7 terms + Sum of next 7 terms
S14 = 63 + 161 = 224
By using the formula of the sum of n terms of an A.P.
Sn = n[2a + (n − 1)d] / 2.
So, S7 = 7(2a + (7 − 1)d) / 2
=> 7(2a + 6d) / 2 = 63
=> 2a + 6d = 18 . . . . (1)
Also, S14 = 14(2a + (14 − 1)d) / 2
=> 14(2a+13d)/2 = 224
=> 2a+13d = 32 . . . . (2)
Now, subtracting eq(1) from eq(2), we get
=> 13d – 6d = 32 – 18
=> 7d = 14
=> d = 2
On putting d = 2 in eq(1), we get,
=> 2a + 6(2) = 18
=> 2a = 18 – 12
=> a = 3
Thus, a28 = a + (28 – 1)d = 3 + 27 (2) = 3 + 54 = 57
Hence, the 28th term is 57.
This is the video solution for this question. Thank you!!