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The 3rd and 6th terms of a series in A.P. are 13 and 31 respectively. Find 20th term.

Q. The 3rd and 6th terms of a series in A.P. are 13 and 31 respectively. Find 20th term.

Solution:

Let us assume that first term=a, common difference=d.

Given,

3rd term=t₃=13

⇒a+(3-1)d=13 (by the formula tₙ=a+(n-1)d)

⇒a+2d=13……………..(i)

and,

6th term=t₆=31

⇒a+(6-1)d=31

⇒a+5d=31………………(ii)

(ii)-(i)⇒(a+5d)-(a+2d)=31-13

⇒3d=18

⇒d=18/3

⇒d=6

Now substituting the value of d to the equation (i)

(i)⇒a+2*6=13

⇒a+12=13

⇒a=13-12

⇒a=1

So, 20th term=t₂₀=a+(20-1)d

=1+19*6

=1+114

=115

Find the three numbers in geometric progression(G.P.) whose sum is 26 and product is 216.

Find the three numbers in geometric progression(G.P.) whose sum is 26 and product is 216.

Solution:

Let us assume that the three numbers in geometric progression are a/r,a and ar.
Given,product is 216.
So, (a/r)*a*ar=216
⇒a³ =216
⇒a³=6³
⇒a=6
Also given that
(a/r)+a+ ar=26
⇒(6/r)+6+6r=26
⇒(6/r)+6r =26-6
⇒(6/r)+6r=20
⇒(6+6r²)/r=20  (by taking the LCM)
⇒6+6r²=20r
⇒3+3r²=10r  (dividing the equation by 2)
⇒3r²-10r+3=0
⇒3r²-(9+1)r+3=0
⇒3r²-9r-r+3=0
⇒3r(r-3)-(r-3)=0
⇒(r-3)(3r-1)=0
either r-3=0⇒r=3
or, 3r-1=0⇒3r=1⇒r=1/3
if r=3 then the numbers are: 6/3,6, 6*3 ie 2,6,18.
If r=1/3 then the numbers are: 6/(1/3),6,6*(1/3) ie 6*3,6,6/3 ie 18,6,2.

A man saved Rs. 16,500 in 10 years. In each year after the first he saved Rs. 100 more than he did in the preceding year. How much did he save in the first year?

A man saved Rs. 16,500 in 10 years. In each year after the first he saved Rs. 100 more than he did in the preceding year. How much did he save in the first year?
Solution:
Let us assume that the amount the man saved in the  first year=a
Here S=16,500
          n=10
          d=100
We are to find the value of a.
We know,
Sₙ= (n/2)*{2a+(n-1)*d}
⇒16,500=(10/2)*{2a+(10-1)*100}
⇒16,500=5*{2a+9*100}
⇒16,500/5=2a+900
⇒3300=2a+900
⇒3300-900=2a
⇒2400=2a
⇒2400/2=a
⇒1200=a
⇒a=1200.
Therefore the man saved in the first year=Rs. 1200