Question

The value of K, for which the term
7 K+3, 4K-5 , 2K+10
are in AP is.​

Answer 1

Given

7K + 3

4K - 5

2K + 10

are in arithmetic progression

=> Common difference is going to be the same

$\implies 4k-5-7k-3 = 2k+10-4k+5$

$\implies -3k - 8 = -2k + 15$

$\implies -3k + 2k = 15+8$

$\implies -k = 23$

$\implies k = -23$

$\boxed{\boxed{\bold{Therefore, \ the \ value \ of \ k \ is \ -23}}}}}$

                                                     

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

The value of K, for which the term

7 K+3, 4K-5 , 2K+10

are in AP is.

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• the term 7K+3, 4K-5 , 2K+10

are in AP is.

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Value of k

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

We know,

The difference between two term is common in an A.P.

So , if p,q,r in A.P.

Then,

• q - p = r - q

Here,

( 7K+3, 4K-5 , 2K+10 are in A.P. )

➠ ( 4k - 5) - (7k+3) = (2k+10)-(4k-5)

➠ -3k -8 = -2k + 15

➠ -3k + 2k = 15 + 8

➠ -k = 23

➠ k = -23

$\Large{\underline{\mathfrak{\bf{Thus}}}}$

• Value of k = -23

$\Large{\underline{\underline{\mathfrak{\bf{Verification}}}}}$

( 7K+3, 4K-5 , 2K+10 are in A.P. )

Keep value of k ,

so, term will be,

➠7×(-23)+3, 4×(-23)-5,2×(-23)+10 are in A.P.

➠ -161+3 , -92-5 , -46+10 are in A.P.

➠ -158 , -97 , -36 are in A.P.

If, they are in A.P.

So, there common difference between two term always same .

➠ -97-(-158) = -36 - (-97)

➠ - 97 + 158 = -36 + 97

➠ 61 = 61

L.H.S.=R.H.S.

That's proved.