1. ARITHMETIC PROGRESSION
EX. NO. 5

2. 1. Find four consecutive terms in an
A.P. whose sum is 12 and the sum of
3rd and 4th term is 14.
Sol. Let the four consecutive terms in
an A.P. be a – 3d, a – d, a + d, a + 3d
As per the first condition,
a – 3d + a – d + a + d + a + 3d = 12
 4a = 12
a=3

3. As per the second condition,
a + d + a + 3d = 14
2a + 4d = 14
 2 (3) + 4d = 14
 6 + 4d = 14
 4d = 14 – 6
 4d = 8
d=2

4.  a – 3d = 3 – 3 (2) = 3 – 6 = – 3
a–d=3–2=1
a+d=3+2=5
a + 3d = 3 + 3 (2) = 9
 The four consecutive terms
of A.P. are – 3, 1, 5 and 9.

5. 2. Find four consecutive terms in an A.P.
whose sum is –54 and the sum of
1 and 3 term is – 30.
Sol. Let the four consecutive terms is an
A.P. be a – 3d, a – d, a + d and a + 3d
As per first condition,
a – 3d + a – d + a + d + a + 3d = – 54
∴ 4a = – 54
∴ a = –54/4
∴ a = - 13.5
st
rd

6. As per the second condition,
∴ a – 3d + a + d = – 30
∴ 2a – 2d = – 30
∴ 2(a – d) = - 30
∴ a – d = -30 / 2
∴ a – d = - 15
∴ - 13.5 - d = - 15
∴ - d = - 15 + 13.5
∴ - d = - 1.5
∴ d = 1.5

7. The four consecutive terms are
a – 3d = - 13.5 – 3(1.5) = - 18
a – d = -13.5 – 1.5 = - 15
a+d = - 13.5 + 1.5 = - 12
a + 3d = - 13.5 + 3 (1.5) = - 9
The four consecutive terms of an
A.P. are – 18, – 15, – 12 and – 9.