Question

The first,
second and last terms of an A.P,
series are 5,9 and 101 resp, find the
mumber of term A.P serices and also
find
find the sum of
all term​

Answer 1

Answer:

$\bigstar{\bold{Number\:of\:terms=25}}$

$\bigstar{\bold{Sum=1325}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• First term = 5
• Second term = 9
• Last term = 101

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Number of terms
• Sum of the terms

$\Large{\underline{\underline{\bf{Solution:}}}}$

➤ First we have to find the number of terms of the A.P

➤The nth term of an A.P is given by

  $a_n=a_1+(n-1)\times d$

where the  first term is 5, last term is 101 and d = 9 - 5 = 4

➤ Substituting the datas we get,

  101 = 5 + ( n-1) × 4

    101 = 5 + 4n - 4

    101 = 4n + 1

    4n = 101 - 1

    4n = 100

      n = 100/4

      n = 25

➤ Hence number of terms of the A.P is 25.

$\boxed{\bold{Number\:of\:terms=25}}$

➤ Now we have to find the sum of the 25 terms of the A.P

➤ Sum of 25 terms of the A.P is given by

    $S_{25}=\dfrac{n}{2}(a_1+a_n)$

➤ Substitute the datas,

    Sum = 25/2 ( 5 + 101)

    Sum = 25/2 × 106

    Sum = 25 × 53

    Sum = 1325

➤ Hence the sum of the terms is 1325

$\boxed{\bold{Sum=1325}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

➤ The last term of an A.P is given by

    $a_n=a_1+(n-1)\times d$

➤ Sum of n terms of an A.P is given by

    $S_n=\dfrac{n}{2}(a_1+a_n)$

   $S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)$