Question

O 5
3) In an AP. 19th term is 52 and 3 points
38th term is 128. Find the sum
of the first 56 terms. *​

Answer 1

Answer:-

Given:

19th term of an AP = 52

38th term = 128

We know that,

nth term of an AP – a(n) = a + (n - 1)d

Hence,

→ a + (19 - 1)d = 52

→ a + 18d = 52 -- equation (1)

Similarly,

→ a + (38 - 1)d = 128

→ a + 37d = 128 -- equation (2)

Subtract equation (1) from (2).

→ a + 37d - (a + 18d) = 128 - 52

→ a + 37d - a - 18d = 76

→ 19d = 76

→ d = 76/19

→ d = 4

Substitute the value of d in equation (1).

→ a + 18 * 4 = 52

→ a = 52 - 72

→ a = - 20

We know,

Sum of first n terms of an AP – S(n) = n/2 * [ 2a + (n - 1)d ]

Hence,

→ S(56) = 56/2 * [ 2 (- 20) + (56 - 1) * (4) ]

→ S(56) = 28 * [ - 40 + 55 * 4 ]

→ S(56) = 28 * [ - 40 + 220 ]

→ S(56) = 28 * (180)

→ S(56) = 5040

Hence, the sum of first 56 terms of the given AP is 5040.

Answer 2

Answer:

Let the First Term of AP be a and Common Difference be d.

↝ tₙ = a + (n - 1)d

• 19th Term of AP :

⇒ t₁₉ = 52

⇒ a + (19 - 1)d = 52

⇒ a + 18d = 52⠀ …( I )

• 38th Term of the AP :

⇒ t₃₈ = 128

⇒ a + (38 - 1)d = 128

⇒ a + 37d = 128⠀ …( II )

• Adding Equation ( I ) & Equation ( II ) :

⇢ (a + 18d) + (a + 37d) = 52 + 128

⇢ 2a + 55d = 180⠀⠀...( III )

━━━━━━━━━━━━━━━━━

• Sum of First 56th Term :

⟶ Sₙ = n/2 [ 2a + (n - 1)d ]

⟶ S₅₆ = 56/2 [ 2a + (56 - 1)d ]

⟶ S₅₆ = 28 [ 2a + 55d ]

• Putting values from Equation ( III )

⟶ S₅₆ = 28 × 180

⟶ S₅₆ = 5040

∴ Sum of First 56 terms is 5040.