Question

Topic - Arthematic Progression
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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

Answer:

$\large{\purple{\ddot{\Mathsdude}}}$

☆●Answered by Rohith kumar maths dude: -

☆In this question given that

☆The sum of 5th and 7th term is 52.

☆Hence,

a+4d+a+6d=52 (From given)

=2a+10d=52 (i)

▪And here also given 10th term is 46.

▪We can written it as ,

a+9d=46. (ii)

☆After multiplying the equation by 2 in ii) equation we get,

2a+18d=92 (ii) .

☆Solving the equation i) and ii)

☆we get,

8d=40

d=40/8

d=5.

☆And substituting the value of d in i) equation

we get,

2a+(10×5)=52

2a=52-50

2a=2

a=2/2

a=1.

▪Hence , the required Ap"s are

●1,6,11,16,21,26,31, ....etc.

☆Hope it helps u mate

☆Thank you.