Question

If the ratio of the fifth and tenth term of an A.P is 1:2 then find the ratio of the first term to the common difference of the given A.P?​

Answer 1

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The ratio of the fifth and tenth terms of an AP is 1 : 2.

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The ratio of the first term to the common difference of the given A.P

$\bf{\underline{\underline\green{Solution:-}}}$

As we know that

The nth term of an AP are given by the formula:-

$\boxed{ \rm \leadsto a_{n} = a + (n - 1)d}$

Here:-

• a = first term

• n = number of terms

• d = common difference

Now

The 5th term

${ \rm \implies a_{5} }$

${ \rm \implies a + (5 - 1)d}$

${ \rm \implies a + 4d}$

The 5th term = a + 4d

The 10th term

${ \rm \implies a_{10} }$

${ \rm \implies a + (10 - 1)d}$

${ \rm \implies a + 9d}$

The 10th term = a + 9d

The ratio of the fifth and tenth terms of an AP is 1 : 2.

${ \rm \implies \dfrac{ a_{5}}{ a_{10}} = \dfrac{1}{2}}$

${ \rm \implies \dfrac{ a + 4d}{ a + 9d} = \dfrac{1}{2}}$

By cross multiplication

${ \rm \implies { 2(a + 4d)} = { 1(a + 9d)}}$

${ \rm \implies { 2a + 8d} = { a + 9d}}$

${ \rm \implies { 2a -a} = { 9d - 8d}}$

${ \rm \implies { a} = { d}}$

${ \rm \implies { \dfrac{a}{d} } = { \dfrac{1}{1} }}$

${ \rm \implies { a :d } = { 1 : 1 }}$

$\underline{ \boxed{{ \purple{\rm The \: ratio \: of \: first \: term \: to\:the \: common \: difference \: = 1 : 1}}} }$