Question

determine the sum of first 35 terms of A.P , if a2=2 and a7=22​

Answer 1

Given :-

2nd term a2 = 2 and 7th term = 22

Solution :-

[ An arithmetic Progression is a list of numbers in which each term obtained by adding a fixed number to the preceeding term except the first term ]

a = First term of AP

d = difference

n = No. of terms

Now , Let's solve the question,

Here,

2nd term of an AP a2 = a + d = 2 ...eq( 1 )

7th term of an AP a7 = a + 6d = 22 eq( 2)

Now, Using Elimination method,

a + 6d = 22

a + d = 2

5d = 20

d = 20/5 = 4

Thus, The value of d = 4

Now,

Put the value of d in eq( 1 )

a + d = 2

a + 4 = 2

a = 2 - 4

a = -2

Therefore, The value of a = -2

Now,

By using formula,

S = n/2 [ 2a + ( n - 1 )d ]

Here, a = -2 , d = 4 , n = 35

Put the required values in the formula,

S = 35/2 [ 2 * -2 + ( 35 - 1 )4

S = 35/2 [ -4 + 136 ]

S = 35/2 * 132

S = 35 * 66

S = 2,310

Hence, The sum of 35 terms of an AP is 2310 $.$

Formula kept in mind :-

• an = a + ( n - 1 )d

• S = n/2 + [ 2a + (n - 1)d ]

Answer 2

$\large\underline{\boxed{\bold\red{Question}}}$

Determine the sum of first 35 terms of an A.P, if the second term is 2 and the seventh-term is 22.

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$\large\underline{\boxed{\bold\blue{Answer}}}$

$\large\underline{\boxed{\bold\purple{Given}}}$

$\large\pink{a2 = 2}$

$\large\pink{a7 = 22}$

$\large\pink{n= 35}$

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$\large\orange{a2 = a + d = 2...(1)}$

$\large\orange{a7 = a6 + d...(2)}$

$\large\orange{substituting \: (1) \: and \: (2), \: we \: get}$

$\orange{\dfrac{\cancel{a + d}}{\cancel{ - a - 6d}}=\dfrac{\cancel2}{ \cancel{- 22}}}$

$\orange{- 5d= - 20}$

$\orange{d=\dfrac{ - 5}{ - 20}}$

$\large\blue{\boxed{d=4}}$

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$\large\orange{a2 = a+d}$

$\large\orange{-a = d - a2}$

$\large\orange{-a = 4-2}$

$\large\orange{-a= 2}$

$\large\orange{a = -2}$

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Sum of first 35th term

$\red{Sn = \dfrac{n}{2}[2a+(n-1)d]}$

$\orange{S35 = \dfrac{35}{2}[2(-2)+(35 - 1) \times 4]}$

$\orange{S35 = \dfrac{35}{\cancel2}[\cancel{-4}+34×\cancel4]}$

$\large\orange{S35 = 35[-2+34×2]}$

$\large\orange{S35 = 35[ - 2 + 68]}$

$\large\orange{S35 = 35[ 66]}$

$\large\blue{\boxed{S35 = 2310}}$

[Hope this helps you.../]