Arithmetic and geometric progressions 53

Hence

35 × 2

7

= 2 a+ 7. 2

10 = 2 a+ 7. 2

Thus 2 a= 10 − 7. 2 = 2 .8,

from which a=

2. 8

2

= 1. 4

i.e.the first term,a=1.4

Problem 7. Three numbers are in arithmetic

progression. Their sum is 15 and their product is 80.

Determine the three numbers.

Let the three numbers be(a−d),aand(a+d)

Then (a−d)+a+(a+d)=15, i.e. 3a=15, from
which,a= 5

Also,a(a−d)(a+d)=80, i.e.a(a^2 −d^2 )= 80

Since a= 5 , 5 ( 52 −d^2 )= 80

125 − 5 d^2 = 80

125 − 80 = 5 d^2

45 = 5 d^2

from which,d^2 =

45

5

=9. Henced=

√

9 =±3.

The three numbers are thus( 5 − 3 ),5and( 5 + 3 ),i.e.
2, 5 and 8.

Problem 8. Find the sum of all the numbers

between 0 and 207 which are exactly divisible by 3.

Theseries3,6,9,12,...,207 is an AP whose first term
a=3andcommondifferenced= 3

The last term is a+(n− 1 )d= 207

i.e. 3 +(n− 1 ) 3 = 207 ,

from which (n− 1 )=

207 − 3

3

= 68

Hence n= 68 + 1 = 69

The sum of all 69 terms is given by

S 69 =

n

2

[2a+(n− 1 )d]

=

69

2

[2( 3 )+( 69 − 1 )3]

=

69

2

[6+204]=

69

2

( 210 )= 7245

Problem 9. The first, twelfth and last term of an

arithmetic progression are 4, 31^12 , and 376^12

respectively. Determine (a) the number of terms in

the series, (b) the sum of all the terms and (c) the

‘80’th term.

(a) Let the AP bea,a+d,a+ 2 d,...,a+(n− 1 )d,

wherea= 4

The 12th term is:a+( 12 − 1 )d= (^3112)
i.e. 4 + 11 d= 3112 ,
from which, 11d= 3112 − 4 = (^2712)
Henced=
(^2712)
11
= (^212)
The last term isa+(n− 1 )d
i.e. 4+(n− 1 )
(
(^212)
)
= (^37612)
(n− 1 )=
37612 − 4
(^212)

(^37212)
(^212)
= 149
Hence the number of terms in the series,
n= 149 + 1 = 150
(b) Sum of all the terms,
S 150 =
n
2
[2a+(n− 1 )d]

150
2
[
2 ( 4 )+( 150 − 1 )
(
2
1
2
)]
= 75
[
8 +( 149 )
(
2
1
2
)]
=85[8+ 372 .5]
= 75 ( 380. 5 )= 28537
1
2
(c) The 80th term is:
a+(n− 1 )d= 4 +( 80 − 1 )
(
(^212)
)
= 4 +( 79 )
(
(^212)
)
= 4 + 197. 5 = (^20112)