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The 26th , 11th and the last term of an A. P are 0,3 and -1/5 respectively. Find the common difference and the number of terms..

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Answer 1

Answer:

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{♣︎given}}}$

• The 26th , 11th and the last term of an A. P are 0,3 and -1/5 respectively. Find the common difference and the number of terms..

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{♣︎to \: find}}}$

a26 = 0 , a11 = 3 and an (last term) = -1/5 of an A.P.

We know that, nth term an = a + (n – 1)d

Then,

a26 = a + (26 – 1)d

⇒ a + 25d = 0

And,

a11 = a + (11 – 1)d

⇒ a + 10d = 3

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{♣︎solving}}}$

⇒ a + 25d – (a + 10d) = 0 – 3

15d = -3

⇒ d = -1/5

we get a + 25(-1/5) = 0 a = 5

Now, given that the last term an = -1/5

⇒ 5 + (n – 1)(-1/5) = -1/5

5 + -n/5 + 1/5 = -1/5

25 – n + 1 = -1

n = 27

Therefore, the A.P has 27 terms and its common difference is -1/5.