Go In an AP. the first term is -5 and the last
form is 45. If the sum of all numbers in the
A-
Pjs 120, then bow ma

1 thought on “Go In an AP. the first term is -5 and the last<br />form is 45. If the sum of all numbers in the<br />A-<br />Pjs 120, then bow ma”

1. Answer:

   Given:

   1. First term, a= -5
   2. Last term, [tex]a_{n}[/tex] = 45
   3. Sum of all the terms, [tex]S_{n}[/tex]= 120

   To find : The number of terms

   Proof:

   Let the number of terms= n

   According to the question,

   [tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] ( a+ [tex]a_{n}[/tex] )

   Substituting the values, we get:

   = 120 = [tex]\frac{n}{2}[/tex] ( -5 +45 )

   ⇒ 120 = [tex]\frac{n}{2}[/tex] (40)

   ⇒ 120 = n (20)

   ⇒ n = [tex]\frac{120}{20}[/tex]

   ⇒ n = 6

   ∴ Number of terms in the given A.P. = n = 6

   Now,

   [tex]a_{n}[/tex] = 45 (Given)

   But,

   [tex]a_{n}[/tex] = a +(n-1) d

   Where

   1. a= first term
   2. [tex]a_{n}[/tex] = last term
   3. n= total number of terms
   4. d = common difference.

   Substituting the values in the equation, we get:

   = 45 = -5 + (6-1) d

   ⇒ 45 +5 = 5d

   ⇒ 50 = 5d

   ⇒ d= [tex]\frac{50}{5}[/tex]

   ⇒ d = 10

   ∴ The common difference = d= 10

   Hence, number of terms = n= 6

   and the common difference = d = 10

   Proved.

   Hope you got that.

   Thank You.

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