Question 3 (iv)
In an AP:
(iv) Given $a_{3} = 15, S_{10} = 125$ , find d and $a_{10}$ .

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Solution

(iv) $G i v e n t h a t, a_{3} = 15, S_{10} = 125 A s a_{n} = a + (n - 1) d, a_{3} = a + (3 - 1) d 15 = a + 2 d \dots \dots (i) S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{10} = \frac{10}{2} [2 a + (10 - 1) d] 125 = 5 (2 a + 9 d) 25 = 2 a + 9 d \dots \dots (i i) O n m u l t i p l y i n g e q u a t i o n (i) b y 2, w e g e t, 30 = 2 a + 4 d \dots \dots (i i i) O n s u b t r a c t i n g e q u a t i o n (i i i) f r o m (i i), w e g e t, - 5 = 5 d d = - 1 F r o m e q u a t i o n (i), 15 = a + 2 (- 1) 15 = a - 2 a = 17 a_{10} = a + (10 - 1) d a_{10} = 17 + (9) (- 1) a_{10} = 17 - 9 = 8$

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Similar questions

Question 3 (iv)
In an AP:
(iv) Given $a_{3} = 15, S_{10} = 125$ , find d and $a_{10}$ .

a_{3} = 15, S_{10} = 125

d

a_{10}

a_{3} = 15, S_{10} = 125

a_{10}

Given , $a_{3} = 15, s_{10} = 125$ find $d$ and $a_{10}$ .

a_{3} = 15, S_{10} = 125

a_{10}

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