Question

The 26th,11th and last term of an A. P. are 0,3 and -1/5, respectively. find the common difference and the number of the terms.​

Answer 1

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▪ Given :-

For an A.P.

• 26th Term = 0

• 11th Term = 3

• Last Term = $\sf-\dfrac{1}{5}$

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▪ To Find :-

• Common Difference

• Number of Terms

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▪ Formula For nth Term :-

$\large \mathtt{a_n= a + (n - 1)d}$

Where :

• a = First term

• d = Common difference

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▪ Solution -

We Have ,

$\large \mathtt{a_{26} = a + 25d= 0} \: \: - - - (1)\\ \\ \large \mathtt{a_{11} = a + 10d = 3} \: \: - - - (2)$

Let,

pth term be the last term of corresponding A.P.

So,

$\large\mathtt{a_p} = - \dfrac{1}{5}$

Subtracting (2) from (1) We Get ,

$\large \mathtt{15d = - 3} \\ \\ :\large \mathtt{\longmapsto d = - \dfrac{3}{15} } \\ \\ \Large \mathtt{ : \longmapsto \pink{ \boxed{ \boxed{ \bf d = - \frac{1}{5} }}}}$

Putting Value of d in (1) We Get,

$\large \mathtt{a + 25 \bigg( - \dfrac{1}{5} } \bigg) = 0 \\ \\ \large \mathtt{ : \longmapsto a - 5 = 0} \\ \\ \LARGE \mathtt{ : \longmapsto \pink{ \boxed{ \boxed{ \bf a = 5}}}}$

As Last term is pth term

Hence,

$\large \mathtt{a_p = a + (p - 1)d} = - \dfrac{ 1}{5} \\ \\ \large \mathtt{ : \longmapsto5 + (p - 1)( - \frac{1}{5} )} = - \frac{1}{5} \\ \\ \LARGE \mathtt{ :\longmapsto\pink {\boxed{\boxed{\bf p = 27}}}}$

Hence,

Their are 27 terms in the A.P.

So ,

• Common Difference = $\bf-\dfrac{1}{5}$

• Number of Terms = 27

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Answer:

Given :-

• The 26th, 11th and last term of an AP are 0, 3 and - 1/5 respectively.

To Find :-

• What is the value of common difference and the number of the terms.

Formula Used :-

$\clubsuit$ The nth term of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms of an AP
• d = Common difference of an AP

Solution :-

First, we have to make the equation :

Given :

• Number of terms = 26

$\leadsto \sf a + (26 - 1)d =\: 0$

$\implies \sf\bold{\green{a + 25d =\: 0\: ------\: (Equation\: No\: 1)}}$

Again,

Given :

• Number of terms = 11

$\leadsto \sf a + (11 - 1)d =\: 3$

$\leadsto\sf\bold{\green{a + 10d =\: 3\: ------\: (Equation\: No\: 2)}}$

Again,

$\leadsto \sf \bold{\green{a + (n - 1)d =\: - \dfrac{1}{5}\: ------\: (Equation\: No\: 3)}}$

Now, by subtracting the equation no 1 and 2 we get,

$\implies \sf a + 25d - (a + 10d) =\: 0 - 3$

$\implies \sf a + 25d - a - 10d =\: - 3$

$\implies \sf {\cancel{a}} {\cancel{- a}} + 25d - 10d =\: - 3$

$\implies \sf 15d =\: - 3$

$\implies \sf d =\: - \dfrac{\cancel{3}}{\cancel{15}}$

$\implies \sf \bold{\red{d =\: - \dfrac{1}{5}}}$

Hence, the common difference of an AP is - ⅕.

Again, by putting the value of d in the equation no 2 we get,

$\implies \sf a + 10d =\: 3$

$\implies \sf a + 10\bigg(- \dfrac{1}{5}\bigg) =\: 3$

$\implies \sf a + \bigg(- \dfrac{10}{5}\bigg) =\: 3$

$\implies \sf a - \dfrac{10}{5} =\: 3$

$\implies \sf a =\: 3 + \dfrac{10}{5}$

$\implies \sf a =\: \dfrac{15 + 10}{5}$

$\implies \sf a =\: \dfrac{\cancel{25}}{\cancel{5}}$

$\implies \sf\bold{\purple{a =\: 5}}$

Again, by putting the value of a and d in the equation no 3 we get,

$\longrightarrow \sf a + (n - 1)d =\: - \dfrac{1}{5}$

$\longrightarrow \sf 5 + \bigg(n - 1\bigg)\bigg(- \dfrac{1}{5}\bigg) =\: - \dfrac{1}{5}$

$\longrightarrow \sf \bigg(n - 1\bigg)\bigg(- \dfrac{1}{5}\bigg) =\: - \dfrac{1}{5} - 5$

$\longrightarrow \sf \bigg(n - 1\bigg)\bigg(- \dfrac{1}{5}\bigg) =\: \dfrac{- 1 - 25}{5}$

$\longrightarrow \sf \bigg(n - 1\bigg)\bigg(- \dfrac{1}{5}\bigg) =\: \dfrac{- 26}{5}$

$\longrightarrow \sf \bigg(n - 1\bigg)\bigg(- \dfrac{1}{5}\bigg) =\: \dfrac{- 26}{5}$

$\longrightarrow \sf (n - 1) =\: - \dfrac{26}{5} \times \bigg(- \dfrac{5}{1}\bigg)$

$\longrightarrow \sf n - 1 =\: \dfrac{26}{5} \times \dfrac{5}{1}$

$\longrightarrow \sf n - 1 =\: \dfrac{130}{5}$

$\longrightarrow \sf n - 1 =\: 26$

$\longrightarrow \sf n =\: 26 + 1$

$\longrightarrow \sf\bold{\red{n =\: 27}}$

${\small{\bold{\underline{\therefore\: The\: common\: difference\: is\: - \dfrac{1}{5}\: and\: the\: number\: of\: terms\: is\: 27\: .}}}}$