Question

In an increasing GP,the sum of the first and last term is 66,the product of the second and the last but one is 128 and the sum of the terms is 126.how many terms are there in progression

Answer 1

5is the and yes and best

Answer 2

Answer:

There are a total of six terms in this (Geometric Progression) GP.

Step-by-step explanation:

Given the GP is an increasing GP.

The sum of the first term of the GP and the last term of the GP is 66.

$a +ar^{n-1}=66$

we will take it as equation (i).

Now the product of the second term of the GP and the second last term of the GP is 128.

$ar \times ar^{n-2}= 128\\\\a^2r^{n-1} = 128\\\\r^{n-1} = \frac{128}{a^2}$

Putting it in the first equation -

$a + ar^{n-1} = 66\\\\a+a\times \frac{128}{a^2} = 66\\\\a^2-66a+128=0\\\\a = 2 \ or\ a =64$

For a = 64,

$r^{n-1} = \frac{128}{64^2} \\\\r^{n-1} = \frac{1}{32}$

But given the Geometric Progression (GP) is increasing so we will reject this case.

For a = 2,

$r^{n-1} = \frac{128}{2^2} \\\\r^{n-1} = 32$

Now, this is acceptable since r>1.

also,

$r^{n-1} = 32\\\\r^n = 32r$

Given the sum of terms of the GP is 126.

$S_n = \frac{a(r^n-1)}{r-1}\\\\126 = \frac{2\times(32r-1)}{r-1} \\\\126(r-1) = 2(32r-1)\\\\126r-126-64r+2=0\\\\62r=124\\\\r=2$

Now we got r = 2,

and we know,

$r^{n-1} = 32\\\\2^{n-1} = 2^5\\\\n-1 = 5\\\\n = 6$

So there are a total of six terms in this Geometric Progression GP.

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