Topic - Arthematic Progression@ Moderators @ Stars @ Best Users The su

1.	Topic - Arthematic Progression@ Moderators @ Stars @ Best Users 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» ♣ Given :- For an A.P. 5th TERM + 7th Term = 52 10th Term = 46 ----------------------- ♣ To FIND :- The Corresponding A.P. ----------------------- ♣ Formula For nth Term :- $\large \mathtt{a_n= a + (n - 1)d}$ Where : a = First term d = Common difference ----------------------- ♣ 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 : $\pink{\huge \mathfrak{6 , 10,14,\: . \: . \: .}}$ $\LARGE\red{\mathfrak{ \text{W}hich \:\:is\:\: the\:\: required} }\\ \Huge \red{\mathfrak{ \text{ A}nswer.}}$ -----------------------

Topic - Arthematic Progression@ Moderators @ Stars @ Best Users 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»

♣ Given :-

For an A.P.

5th TERM + 7th Term = 52

10th Term = 46

-----------------------

♣ To FIND :-

The Corresponding A.P.

-----------------------

♣ Formula For nth Term :-

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

Where :

a = First term

d = Common difference

-----------------------

♣ 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 :

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

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