Question

26th, 11th and last term of an ap are 0, 3 and -1/5 respectively . find the common difference and the number of terms

Answer 1

Given: $26 _{th}$ term of AP series is = 0

           $11_{th\\}$ term of AP series is =  3

          last term of AP series is = -1/5

Find: common difference and number of turns

solution:

as we know that to find $n_{th}$ term we use formula:

                     $a_{n}$ = a₁ + (n-1) d

                     $a_{n}$ = $n^{th}$ term

                     a₁ = first term

                     n = number of terms

                     d = common difference

Step 1:

equation for $11_{th\\}$ term:

              $a_{11}$ = a₁ + (11-1) d

              3 = a₁ + 10 d  ...............(eqn 1)

equation for $26 _{th}$ term:

               $a_{26}$ = a₁ + (26-1) d

               0 = a₁ + 25 d  ...............(eqn 2)

Step 3:

now subtract equation 2 from 1, we get

             d = $\frac{-1}{5}$   .................. (common difference)

put this value of d in equation 1, we get:

             3 = a₁ + 10 d

             3 = a₁ + 10( $\frac{-1}{5}$)

             3 = a₁ + (-2)

              a₁ = 5 ............... (first term)

Step 4:

to find number of terms:

             $a_{n}$ = a₁ + (n-1) d

here take $a_{n}$ as last term, a₁ = first term, d = common difference

so,

              $\frac{-1}{5}$ = 5 + (n-1) ($\frac{-1}{5}$ )

               $\frac{-1}{5}$ = 5 - (n-1)/5

               (n-1)/5 = 5 + $\frac{1}{5}$

               (n-1)/5 = 26/5

               (n-1) = (26/5)*5

               n-1 = 26

               n= 26 + 1

               n = 27 ............... (number of terms)

so the answer is :

common difference = (-1/5)

number of terms = 27

         

Answer 2

Step-by-step explanation:

Given that: 26th, 11th and last term of an ap are 0, 3 and -1/5 respectively .

To find:

find the common difference and the number of terms

Solution:

Let the first term of A.P. is a

common difference is d

no. of terms: n

General term of AP

$\bold{a_n = a + (n - 1)d}$

26th term is 0

$a_{26} = a + 25d \\ \\ a + 25d = 0 \: \: \: ...eq1$

11th term is 3

$a_{11}= a + 10d \\ \\ a + 10d = 3 \: \: \: ...eq2$

From eq1 and eq2

$a + 25d = 0 \\ a + 10d = 3 \\ ( - )( - ) \: \: ( - ) \\ - - - - - - \\ 15d = - 3 \\ \\ d = \frac{ - 3}{15} \\ \\ \bold{d = \frac{ - 1}{5}}$

Put the value of d in eq1

$a + 25d = 0 \\ \\ a + 25 \times ( \frac{ - 1}{5} ) = 0 \\ \\ \bold{a = 5 }$

Now,find the value of n,

as last term is -1/5

$\frac{ - 1}{5} = 5 + (n - 1)( \frac{ - 1}{5} ) \\ \\ \frac{ - 1}{5} - 5 = \frac{ - 1}{5} (n - 1) \\ \\ \frac{ - 26}{5} = \frac{ - 1}{5} (n - 1) \\ \\ 26 = n - 1 \\ \\ \bold{n = 27}$

First term a= 5

Common difference d= -1/5

Total terms= 27

Hope it helps you.