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*Series And Sequence Solved Examples*

*Series And Sequence Solved Examples*

**1.Find the 13th term of G.P. with first term -1 and common ratio 3.**

2

**. Find the common ratio and 8th term of the G.P. 1,-3,9, ...****Solution:**

**3. Find the common ratio of a G.P. whose 5th term is 2 and first term is 1/8.**

**4. Which term of the G.P. is 64/27 whose first term is 27 and the second term is 18?**

**5. The third term of a geometric series is 27 times the 6th term and the 4th term is 9. Find the series.**

**10. The sum of first four terms is a G.P. is 30 and that of the last four terms is 960.If its first term is 2, find the common ratio.**

**Solution:**

Let r be the common ratio then,

2+2r+2r²+2r³ = 30

or, 1+r+r²+r³ = 15

or r³+r²+r+14 = 0

or r³-2r²+3r²-6r+7r -14 = 0

or r²(r-2)+3r(r-2)+7(r-2) = 0

or (r-2) (r²+3r+7)

∴ r=2

**11. Some geometrical means are inserted between 5 and 160. If the third mean between them is 40, find the number of means.**

**Solution:**

Here, the third mean =40

we also know, third mean=4th term

So, ar³=40

or, 5.r³ = 40

or, r³ =8 =2³

∴r = 2

Now,

tₙ=arⁿ⁻¹

or, 160 = 5.2ⁿ⁻¹

or, 32= 2ⁿ⁻¹

or, 2⁵ = 2ⁿ⁻¹

or, n-1 = 5

∴n=6

So, the number of mean =n-2 =4

**12. If A.M. and G.M. between two numbers are 10 and 8 respectively, find the two numbers.**

**Solution:**

Let a and b be the two numbers whose A.M. is 10 and G.M. is 8.

Then, (a+b)/2 = 10

or, a+b = 20 ...(i)

And √ab =8

or, ab =64

or, b = 64/a ...(ii)

Putting the value of b in equation (i) we get

a+64/a =20

or, a²+64 =20a

or, a²-20a+64 =0

or, a²-16a-4a+64 =0

or, a(a-16) -4(a-16) =0

or, (a-16)(a-4) =0

∴a = 16 or 4

when a=16, then b= 64/16 =4

when a = 4, then b =64/4 =16

Hence, required two numbers are 16 and 4 or 4 and 16.