Question

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Answer 1

Question

Determine k , so that (k +2) , (4k-6) and (3k-2) are three consecutive terms of an A.P.

Solution

Given :-

• Three terms of an A.P. , ( k +2) , (4k-6), (3k-2) ..........

Find :-

• Value of K.

Explanation,

We Know,

If x,y,z, are terms of an A.P. series ,

So, All common difference will be same .

As.

• (y-x) = (z-y)

So,

➩ ( 4k-6) - (k+2) = (3k-2) - (4k-6)

➩ (4k-k) + (-6 - 2 ) = (3k-4k)-(2-6)

➩ 3k - 8 = -k + 4

➩ 3k + k = 4 + 8

➩ 4k = 12

➩ k = 12/4

➩ k = 3

Hence

• Value of k will be = 3

________________

Answer 2

$\large\underline\bold{Question:-}$

Determine k so that k+2, 4k-6 and 3k-2 are three consecutive terms of an AP.

$\large\underline\bold{Answer:-}$

$\small\underline\bold{Given:-}$

$\sf\ Three\: consecutive\: terms\: of\: AP = k+2, \: 4k-6, \:and\: 3k-2.$

$\small\underline\bold{To\: Find:-}$

$\sf\ Value\:of\:k$

$\small\underline\bold{Solution:-}$

$\sf\ In\:the\:terms\:of\:an\:AP\: the\: difference\: between$

$\sf\ two\: consecutive\: terms\:must\:be\:same.$

$\sf\ So, \:if\:k+2,\:4k-6\:and\: 3k-2 \: are \: the \: terms\: of\:an\:AP.$

$\sf\ Then,$

⟹$\sf\ 4k-6\: -\: (k+2)= 3k-2(4k-6)$

⟹$\sf\ 4k-6\: -\: k-2=3k-2-4k+6$

⟹$\sf\ 3k-8=\:-k+4$

⟹$\sf\ 4k=12$

⟹$\sf\large\ k= \frac{12}{4}$

$\large\underline\bold\purple{k=3}$

$\sf\ Thus\:the\:value\:of\:k\:is\:3.$