Question

the 6th and and 20 th term of an ap is 8 and-20. find the 30th term?

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Answer 1

Given ,

$\star$a + 5d = 8 --------- eq (i)

$\star$a + 19d = - 20 ----- eq (ii)

Using elimination method , subtract (i) from (ii)

$\sf \implies (a + 19d) - (a + 15d) = -20 - 8 \\ \\ \sf \implies </p><p>14d = -28 \\ \\ \sf \implies </p><p>d = - \frac{28}{14} \\ \\ \sf \implies </p><p>d = -2$

Put the value of d = -2 in eq (i)

$\sf \implies a + 5(-2) = 8 \\ \\ \sf \implies </p><p>a - 10 = 8 \\ \\ \sf \implies </p><p>a = 18$

Therefore ,

$\large \sf \fbox{ \fbox{ a_{30}= a + 29d</p><p> }}$

$\sf= 18 + 29(-2) \\ \\ \sf</p><p>= 18 - 58 \\ \\ \sf</p><p>= 40 </p><p>$

Hence , 40 is the 30th term of a given AP