ARITHMETIC
PROGRESSIONS
10) The 17th term of an AP exceeds its 10th term by 7.
Find the common difference.
Sol:
We need to find d
For given AP:
a17 =
a10 +
7
a17= a + 16d
∴ a + 16d =
a10= a + 9d
a + 9d
+ 7
∴ 16d – 9d =
7
∴ 7d =
7
∴ d =
1
∴ Common difference of AP is 1.
Exercise 5.2 10
HOMEWORK
16) Determine the AP whose third term is 16 and
7th term exceeds the 5th term by 12.
Sol:
We need to determine the AP
a3 = 16,
a7 =
a5 +
12
a3 = a + 2d
∴ 16
= a + 2d
We know that,
…..(i)
a7 = a5 + 12
a7 = a + 6d
∴ a + 6d =
a5 = a + 4d
a + 4d
+ 12
∴ 6d – 4d =
12
∴ 2d =
12
∴ d =
6
Substituting d = 6 in (i)
a + 2(6) = 16
∴ a + 12 = 16
∴ a =
16 – 12
∴ a =
4
That means,
First term of AP is 4 and
common difference is 6
∴ The required AP is 4, 10, 16, 22, ...
a2 = 4 + 6 = 10
a3 = 10 + 6 = 16
a4 = 16 + 6 = 22
Exercise 5.2 16