Question

ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯಲ್ಲಿ 6 ನೇ ಪದವು 3 ನೇ ಪದದ ಎರಡರಷ್ಟಕ್ಕಿಂತ ಒಂದು ಹೆಚ್ಚಾಗಿದೆ. 4 ನೇ ಮತ್ತು 5ನೇ ಪದಗಳ ಮೊತ್ತವು 2 ನೇ ಪದದ ಐದರಷ್ಟಿದೆ. ಈ ಸಮಾಂತರ ಶ್ರೇಢಿಯ 10 ನೇ ಪದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.​

Answer 1

Given:

ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯಲ್ಲಿ 6 ನೇ ಪದವು 3 ನೇ ಪದದ ಎರಡರಷ್ಟಕ್ಕಿಂತ ಒಂದು ಹೆಚ್ಚಾಗಿದೆ. 4 ನೇ ಮತ್ತು 5ನೇ ಪದಗಳ ಮೊತ್ತವು 2 ನೇ ಪದದ ಐದರಷ್ಟಿದೆ. ಈ ಸಮಾಂತರ ಶ್ರೇಢಿಯ 10 ನೇ ಪದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.​

In an AP 6th term is 1 more than twice to the 3rd term. The sum of the 4th and 5th terms is 5 times the 2nd term. Find 10term of an AP​

To find:

10th term of an AP​

Solution:

We know the nth term of an A.P. is given as:

$\boxed{\bold{a_n = a + (n - 1)d}}$

where aₙ = last term, a = first term, n = no. of terms and d = common difference

So, let us consider the terms of the A.P. as follows:

a, a + d, a + 2d, a + 3d . . .

Therefore,

$6th \:term = a_6 = a + (6-1)d = a + 5d$

$3rd \:term = a_3 = a + (3-1)d = a + 2d$

$4th \:term = a_4 = a + (4-1)d = a + 3d$

$5th \:term = a_5 = a + (5-1)d = a + 4d$

$2nd \:term = a_2 = a + (2-1)d = a + d$

$10th \:term = a_1_0 = a + (10-1)d = a + 9d$

As given that in the AP 6th term is 1 more than twice to the 3rd term, so the equation will be,

$a_6 = 1 + 2(a_3)$

$\implies a + 5d = 1 + 2(a + 2d)$

$\implies a + 5d = 1 + 2a + 4d$

$\implies d - a = 1$

$\implies d = 1 + a$ . . . . (1)

The sum of the 4th and 5th terms is 5 times the 2nd term

$a_4 + a_5 = 5a_2$

$\implies a+3d + a + 4d = 5 (a + d)$

$\implies 2a + 7d = 5a + 5d$

$\implies 2d = 3a$  

On substituting from (1), we get

$\implies 2(1 + a) = 3a$

$\implies 2 + 2a = 3a$

$\implies a = 2$

On substituting the value of a = 2 in equation (1), we get

$d = 1+ a = 1 + 2 = 3$

Therefore, the A.P. will be as follows:

a = 2

a + d = 2+3 = 5

a + 2d = 2 + 2(3) = 2 + 6 = 8

a + 3d = 2 + 3(3) = 2 + 9 = 11

→ 2, 5, 8, 11, . . .  ←

Now,

The 10th term will be,

= a + 9d

= 2 + 9(3)

= 2 + 27

= 29

Thus, the 10th term of the given A.P. is → 29.

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