Question

$\orange{Please- answer}$

In an ap the sum of its first ten terms is 150 and the sum of its next term is 550.
Find the ap​

Answer 1

$let \: the \: first \: term \: of \: ap \: be \: a \: and \: common \: difference \: be \: d. \\ now \\ sum \: of \: first \: 10 \: term = 150 \\ \frac{10}{2} (2a + (10 - 1)d) = 150 \\ 5(2a + 9d) = 150 \\ 2a + 9d = 30 \\ \\ again \: given \: that \: sum \: of \: next \: 10 \: term \: is \: 550 \: therefore \\ \\ sum \: of \: first \: 20 \: terms = 150 + 550 \\ \frac{20}{2} (2a + (20 - 1)d) = 700 \\ 10(2a + 19d) = 700 \\ 2a + 19d = 70 \\ \\ subtracting \: both \: equation \: we \: have \\ 10d = 40 \\ d = 4 \\ \\ putting \: d = 4 \: in \: 2a + 9d = 30 \\ 2a + 36 = 30 \\ 2a = 30 - 36 \\ 2a = - 6 \\ a = \frac{ - 6}{2} \\ a = - 3 \: \: and \: \: d = 4$

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Answer 2

Given:-

Sum of first ten terms of A.P = 150

Sum of its next ten terms = 550.

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Solution:-

Let the First term of A.P = a

And the common difference = d

According to the question,

$\: \: \: \: \: \: \: \: \frac{10}{2} (2a + 9d) = 150 \\ = > 5(2a + 9d) = 150 \\ = > (2a + 9d) = \frac{150}{5} \\ = > (2a + 9d) = 30 \\ = > 2a = (30 - 9d) ......(i)$

Also,

•Sum of first twenty terms = Sum of first ten terms + sum of next ten terms.

$= > \frac{20}{2} (2a + 19d) = 150 + 550 \\ = > 10 (2a + 19d) = 700 \\ = > 2a + 19d = \frac{700}{10}$

Put the value of 2a from eq(i)

$= > 30 - 9d + 19d = 70 \\ = > 10d = 70 - 30 \\ = > d = \frac{40}{10} \\ = > d = 4$

Now, Put the value of d in eq.(i)

$\: \: \: \: \: \: \: \: 2a = 30 - 9 \times 4 \\ = > 2a = 30 - 36 \\ = > a = \frac{ - 6}{2} \\ = > a = - 3$

Therefore, the A.P. is = -3, 1, 5, 9, ......

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