Question

if the 10th term of an AP is twice the 4th term and 23rd term is K times the 8th term then the value OK K is???​

Answer 1

If the 10th term of an AP is twice the 4th term, then the value of k is 2.5

Solution:

Given, the 10th term of an AP is twice the 4th term  and 23rd term is K times the 8th term  

Then we have to find the value of K

$\text { In Arithmetic progression, } t_{n}=a+(n-1) d$  ⇒ (1)

Where “d” is the common difference between the terms

“a” is the first term of sequence

“n” is the nth term of sequence

Given that 10th term of an AP is twice the 4th term.

Hence we can frame a equation as,

$\mathrm{t}_{10}=2 \times \mathrm{t}_{4}$

By using eqn 1, we get

$\begin{array}{l}{a+(10-1) d=2(a+(4-1) d)} \\ {a+9 d=2 a+6 d} \\ {9 d-6 d=2 a-a} \\ {a=3 d}\end{array}$

Given that 23rd term is K times the 8th term:

$\begin{array}{l}{\mathrm{t}_{23}=\mathrm{k} \times \mathrm{t}_{8}} \\ {\mathrm{a}+(23-1) \mathrm{d}=\mathrm{k}(\mathrm{a}+(8-1) \mathrm{d})} \\ {\mathrm{a}+22 \mathrm{d}=\mathrm{k}(\mathrm{a}+7 \mathrm{d})}\end{array}$

By substituting "a" = 3d,

$\begin{array}{l}{3 d+22 d=k(3 d+7 d)} \\\\ {25 d=k(10 d)} \\\\ {25=10 k} \\\\ {K=\frac{25}{10}=2.5}\end{array}$

Hence, the value of k is 2.5.

Answer 2

Answer:

simplest method

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