Question

if 3.4, p, q, r, 7.4 are in A.P., then value of r is​

Answer 1

Solution :-

we know that, when a, b and c are in AP

• b = (a + c)/2

since there are three terms between 3.4 and 5.4 .

so,

→ Middle term (q) = (3.4 + 7.4)/2 = 5.4

now, as we can see that, p is the middle term between 3.4 and 5.4 .

so,

→ Middle term (p) = (3.4 + 5.4)/2 = 4.4

then,

• common difference = a2 - a1 = a3 - a2

so,

→ d = p - 3.4 = q - p = 4.4 - 3.4 = 5.4 - 4.4 = 1

therefore,

→ a4 = (a3 + d)

→ r = 5.4 + 1

→ r = 6.4 (Ans.)

Hence , value of r is equal to 6.4 .

Method 2) :-

Let given 5 terms in AP be a-2d, a-d, a, a+d and a+2d .

so,

→ (a - 2d) + (a + 2d) = 3.4 + 7.4

→ 2a = 10.8

→ a = 5.4

now,

→ a - 2d = 3.4

→ 5.4 - 2d = 3.4

→ 2d = 5.4 - 3.4

→ d = 1 .

then,

→ (a + d) = 5.4 + 1 = 6.4 .

Hence, value of r is equal to 6.4 .

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Answer 2

SOLUTION

GIVEN

3.4 , p , q , r , 7.4 are in AP

TO DETERMINE

The value of r

EVALUATION

Here it is given that 3.4 , p , q , r , 7.4 are in AP

So common difference exists

Let common difference = d

First term = a = 3.4

Thud we have

2nd term = p = a + d = 3.4 + d

3rd term = q = a + 2d = 3.4 + 2d

4th term = r = a + 3d = 3.4 + 3d

5th term = 7.4 = a + 4d = 3.4 + 4d

Now from the last equation we get

7.4 = 3.4 + 4d

$\sf{ \implies \: 3.4 + 4d = 7.4}$

$\sf{ \implies \: 4d = 7.4 - 3.4}$

$\sf{ \implies \: 4d = 4}$

$\sf{ \implies \: d = 1}$

Thus we get

r = 3.4 + 3d = 3.4 + 3 = 6.4

FINAL ANSWER

Hence the required value of r = 6.4

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