Question

The sum of 5th term and 7th term of
an A.P is 52 and the 10 th term is 46.Find A.P?
​

Answer 1

♣ Given :-

For an A.P.

5th Term + 7th Term = 52

10th Term = 46

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♣ To Find :-

The Corresponding A.P.

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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 ,

$\large \sf{a_{5} +a_{7} = 52}\\ \\ \large \sf{ = a + 4d + a + 6d= 52}\\ \\ \large \sf2a + 10d \: \: - - - (1)\\ \\ \large \sf{a_{11} = a + 10d = 46}$

♠ Multiplying Both Sides by 2 :

$\large :\longmapsto\sf 2a + 20d = 92\: \: - - - (2)$

♠ Subtracting (1) From (2) , We get :-

$\sf10d = 40 \\ \\ \sf d = \cancel{\frac{40}{10} }\\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf d = 4} }}}$

♠ Putting Value of d in (1) :

$:\longmapsto \sf2a + 10 \times 4 = 52 \\ \\ :\longmapsto \sf 2a = 12 \\ \\ :\longmapsto \sf a = \cancel\dfrac{12}{2} \\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf a = 6} }}}$

We Know ,

A.P. Having First Term a and common difference d is is of the Form :

a , a + d , a + 2d , . . . .

Hence ,

The Required A.P is :

$\pink{\huge \mathfrak{6 , 10,14,\: . \: . \: .}}$

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

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Answer 2

Required A.P must be

$\color{navy}{\LARGE \pmb{6 , 10,14,\: . \: . \: .}}$