Question

if 3+k,18-k,5k+1 are in A.P then find k​

Answer 1

We know that :-

If a,b,c are in A.P then :-

=> 2b = a+c

Now , a = 3+k

b = 18-k

c = 5k+1

=> 2(18-k) = 3+k + 5k+1

=> 36-2k = 4+6k

=> -2k-6k = 4-36

=> -2k-6k = -32

=> -8k = -32

=> k = (32)/8

=> k = 4

Answer : 4

Additional Information :-

$\boxed{ \boxed{\tt \large T_n = a+(n-1)d} } \\ \\ \tt where : \\ \\ \tt \bull T_n = n_{th} \: term \\ \\ \tt \bull n = number \: of \: terms \\ \\ \tt \bull a= first \: term \\ \\ \tt \bull d = common \: difference$

$\boxed{ \boxed{\tt \large s_n = \frac{n}{2} \bigg(2a+(n-1)d \bigg)} }$

Answer 2

Answer:

★Question★

• If 3 + k, 18 - k, 5k + 1 are in A.P, then find k

★Answer★

Given :-

• 3 + k, 18 - k, 5k + 1 are in A.P.

To find :-

• Find the value of k.

Concept used :-

• We know that :-
• If a,b,c are in A.P., then b - a = c - b
• ⟹ 2b = a + c

Solution:-

Let us assume that

• a = 3 + k
• b = 18 - k
• c = 5k + 1

We know that :-

If a,b,c are in A.P., then b - a = c - b

⟹ 2b = a + c

On Substituting the values of a, b and c, we get

⟹ 2(18 - k) = 3 + k + 5k + 1

⟹36 - 2k = 4 + 6k

On transposing

⟹ -2k - 6k = 4 - 36

⟹ - 8k = - 32

⟹ k = 4

$\bf\implies \: Value \: of \: k = 4$