Question

A3=15. S10=125,d=? a10=?
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Answer 1

Answer :-

a = 17

d = -1

Given :-

$a_3 = 15$

$s_{10} = 125$

$a_{10} = 8$

To find :-

The value of d and a.

Solution:-

Let a be the first term and d be the common difference.

A/Q

→$a_3 = 15$

→$a+2d = 15-------1$

→$s_{10} = 125$

→$\dfrac{10}{2}[2a + (10-1)d] = 125$

→$5 (2a + 9d) = 125$

→$10 a + 45d = 125$

→$5(2a + 9d) = 125$

→$2a +9d = \dfrac{125}{5}$

→$2a +9d = 25 ------2$

Multiply eq.1 by 2

→$( a + 2d = 15 ) \times 2$

→$2a +4d = 30$

Subtract equation.1 and eq.2

→$2a +4d-(2a+9d)= 30 -25$

→$2a -2a + 4d-9d = 5$

→$-5d = 5$

→$d = \dfrac{-5}{5}$

→$d = -1$

Put the value of d in eq.1

→a + 2d = 15

→a +2 × -1 = 15

→a = -2 = 15

→a = 15 + 2

→a = 17

hence,

The value of a is 17 and d is -1.

→$a_(10) = a + 9d$

→$a_{10} = 17 + 9 \times -1$

→$a_{10} = 17 -9$

→$a_{10} = 8$

hence,

The value of $a_10$ is 8.

Answer 2

GIVEN:

The third term of an AP = 15

a + 2d = 15 -----(1)

Sum of the first ten terms = 125

Sn = n/2 [2a + (n -1)d]

125 = 10/2 [2a + (10 - 1)d]

125 × 2 = 10[2a + (9)d]

250 = 10[2a + 9d]

250/10 = 2a + 9d

25 = 2a + 9d -----(2)

Solve eq (1) and (2)

a + 2d = 15 × 2

2a + 9d = 25 × 1

------------------------

2a + 4d = 30

2a + 9d = 25

(-)

-------------------

-5d = 5

-------------------

d = 5/-5

d = -1

Common Difference = -1

Substitute d in eq - (1)

a + 2d = 15

a + 2(-1) = 15

a - 2 = 15

=> a = 15 + 2

=> a = 17

First-term = 17

a10 = a + 9d

= 17 + 9(-1)

= 17 - 9

a10 = 8

Therefore, a = 17, d = -1 and a10 = 8.