Question

1
. The 26th, 11th and the last term of an AP are 0, 3 and -
common difference and the number of terms.​

Answer 1

Answer:

Step-by-step explanation:

The 26th, 11th and the last term of an AP are 0, 3 and -1/5

common difference and the number of terms.​

Let d and a are common difference and first term and n  is no of terms

Then

$A_{26}=a+25d=0-------(1)\\\\A_{11}= a+10d=3-------(2)\\\\Subtracting\ (2) from\ (1)\\\\15d=-3\\\\d=-\frac{3}{15}=-\frac15$

Putting value of d in(2) we get

$a+10(-\frac15)=3\\\\a-2=3\\\bf a=5$

Now nth term=-1/5

$\therefore 5+(n-1)(-\frac15)=-\frac15\\\\(n-1)(-\frac15)=-\frac15-5=-\frac{26}{5}\\\\n-1 =\frac{-\frac{26}{5}}{-\frac15} =26\\\\\bf \therefore n=27\\\\Hence \ number \ of \ terms=27$