ARITHMETIC
PROGRESSIONS
= 44
Q.18) The sum of the 4th and 8th terms of an A.P is 24 and the
sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Sol:
a4 + a8
= 24
….. (given)
∴
a + 3d
= 24
∴
2a + 10d
= 24
Dividing throughout by 2
a + 5d
= 12
… (i)
a6 + a10
= 44
… (given)
∴
2a + 14d
= 44
Dividing throughout by 2
a + 7d
= 22
… (ii)
Subtracting (i) from (ii)
a
+
7d
= 22
a
+
5d
= 12
(-)
(-)
(-)
2d
= 10
∴
d
= 5
Substituting d = 5 in (i)
a + 5d
= 12
∴
a + 5(5)
= 12
∴
a + 25
= 12
∴
a
= 12 – 25
a2
=
a + d
a3
=
a + 2d
= –13 + 2(5)
= –13 + 10
The first three terms of AP are –13, –8 and –3
a1, a2, a3
= –13 + 5
= – 8
∴
a =
–13
With the values of a & d lets find a2 & a3.
+ a + 7d
∴
a + 5d
+ a + 9d
= – 3
a4 = a + 3d
a8 = a + 7d
a6 = a + 5d
a10 = a + 9d
Exercise 5.2 18