Question

for an ap if t10=50 and t11=55 find t20​

Answer 1

Step-by-step explanation:

Let the first term be 'a' , and let the common difference be 'd'.

Since, we know that

$nth \: term \: = a+(n - 1)d$

So,

$t10 = a + (10 - 1)d \\ t11 = a + (11 - 1)d$

therefore,

$a + 9d = 50..................(1) \\ a + 10d = 55....................(2)$

now, subtracting equation (2) from equation (1) we get,

$9d - 10d = 50 - 55$

$- d = - 5$

$d = 5$

Hence,

$a + 9d = 50..........(from \: 1)$

$a + (9 \times 5) = 50$

$a = 50 - 45$

$a = 5$

Therefore,

$t20 = 5 + (20 - 1)5$

$t20 = 5 + 19×5$

$So, t20 = 5 + 95 = 100$

$t20 = 100$

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