Question

the value of k for which the terms 7k+3,4K-3,2k+10 are in ap​

Answer 1

Question :

To find the value of k for which the terms 7k+3, 4K-3, 2k+10 are in ap.

Theory :

If a,b,c are in Ap

common difference,d =b-a=c-b

Therefore,

$\sf \: b - a = c - b$

$\implies \bf \: 2b = a + c$

Solution :

If 7k+3, 4K-3 , 2k+10 are in Ap

Therefore,

$\sf 2(4k - 3) = 2k + 10 + 7k + 3$

$\sf \implies8k - 6 = 9k + 13$

$\sf \implies 8k - 9k= 13 + 6$

$\sf \implies - k = 22$

$\bf \implies k = - 22$

Therefore the value of k =-22

______________________

More About Arithmetic Progression:

•Genral term of an Ap

$\sf a_{n} = a + (n - 1)d$

•Sum of first n terms of an AP

$\sf \: s_{n} = \dfrac{n}{2} (2a + (n - 1)d)$

Answer 2

Gívєn :

• To find the value of k for which the terms 7k+3, 4K-3, 2k+10 are in ap.

Tσ fínd :

• The value of k

Sσlutíσn :

• $\mathcal {2(4k - 3) = 2k + 10 + 7k + 3}$
• $\mathcal {\implies{8k - 6 = 9k + 13}}$
• $\mathcal {\implies {8k - 9k= 13 + 6}}$
• $\mathcal{\implies{ - k = 22}}$
• $\mathcal{\implies {k = - 22}}$
• Therefore the value of k =-22

Addítíσnαllч :

• If the initial term of an arithmetic progression is $\sf{\displaystyle a_{1}}$and the common difference of successive members is d, then the nth term of the sequence$\sf({\displaystyle a_{n}})$ is given by: $\sf{\displaystyle \ a_{n}=a_{1}+(n-1)d},$ and in general $\sf{\displaystyle \ a_{n}=a_{m}+(n-m)d}$

$\sf\rule{150}1$

• Genral term of an Ap
• $\sf a_{n} = a + (n - 1)d$
• Sum of first n terms of an AP
• $\sf \: s_{n} = \dfrac{n}{2} (2a + (n - 1)d)$