If the $$10th$$ term of an A.P is $$120$$ and its $$20th$$ term is $$110$$, then the sum of first $$200$$ terms is

$$T10=120=a+9d$$−−−$$1T20=1100=a+19d$$−−−$$21$$ and $$2$$⇒ we get $$a=1200$$,$$d=1200S200=20022200+199200$$ sum of n terms in A.$$P=n22a+n-1d=2012=100.5$$

Arithmetic progression

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$If the 10^{th} term of an A.P is \frac{1}{20} and its 20^{th} term is \frac{1}{10}, then the sum of first 200 terms is$

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Solution

$T_{10} = \frac{1}{20} = a + 9 d - - - (1)$
$\begin{array}{l} T_{20} = \frac{1}{100} = a + 19 d - - - (2) \\ (1) and (2) \Rightarrow we get a = \frac{1}{200}, d = \frac{1}{200} \\ S_{200} = \frac{200}{2} [\frac{2}{200} + \frac{199}{200}] (s u m o f n t e r m s i n A . P = \frac{n}{2} (2 a + (n - 1) d)) \\ = \frac{201}{2} = 100.5 \end{array}$

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Arithmetic progression

Remember concepts with our Masterclasses.

If the 10th term of an A.P is 120 and its 20th term is 110, then the sum of first 200 terms is

T10=120=a+9d−−−1T20=1100=a+19d−−−21 and 2⇒ we get a=1200,d=1200S200=20022200+199200 sum of n terms in A.P=n22a+n-1d=2012=100.5

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$If the 10^{th} term of an A.P is \frac{1}{20} and its 20^{th} term is \frac{1}{10}, then the sum of first 200 terms is$