Question

2.Find the sum of first 51 terms of an AP whose second & third terms are 14 & 18 respectively.
5 points
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Answer 1

To Find :

• We have to find sum of First 51 terms of AP.

Solution :

Second term Of AP = 14

Third term of AP = 18

We know that,

$\large\dag \large { \red{\boxed{\bf{a_n = a + (n - 1) }}}}$

Second term can be written as = a + d

Third term can be written as = a + 2d

a + d = 14 -------1)

a + 2d = 18 --------2)

By elimination method Solving equation 1) and 2)

⠀⠀⠀⠀⠀ a + d = 14

⠀⠀⠀⠀⠀ a + 2d = 18

⠀⠀⠀⠀⠀ -- ⠀--⠀ ⠀--⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀- d = - 4

⠀⠀⠀⠀⠀⠀⠀⠀d = 4

• Substituting value of d in (1)

a + d = 14

a + 4 = 14

a = 14 - 4

a = 10

• Sum of 1st 51 terms

we know that,

$\large\dag \large { \red{\boxed{\bf{S_n = \frac{n}{2}[2a + (n - 1)d]}}}}$

= 51/2[2 × 10 + (51 - 1)4]

= 51/2[20 + 50 × 4]

= 51/2[20 + 200]

= 51/2[220]

= 51 × 110

= 5610

$\large\dag \large { \green{\underline{\bf{Hence }}}}$

$\large\dag \large { \purple{\textbf{Sum of 51 terms of AP is \blue{5610} }}}$

Answer 2

Answer:

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