Question

The sum of first 55 term of an A.P. is 3300. Find 28th term.

Answer 1

Answer:

60

Step-by-step explanation:

Sn = n/2[2a + (n – 1)d]

∴ S55 = 55/2[2a + (55 – 1) d]

∴ 3300 = 55/2[2a + 54d]

∴ 3300 = 55/2 × 2[a + 27d]

∴ 3300 = 55 [a + 27d]

∴ 3300/55 = a + 27d

∴ a + 27d= 60 ......(i)

Now, tn = a + (n – 1) d

∴ t28 = a + (28 – 1) d

∴ t28 = a + 27d

Putting the value of a+ 27d

∴ t28 = 60 [From (i)]

∴ Twenty eighth term of A.P. is 60.

Answer 2

Answer:

• 28th term of A.P. is 60.

Step-by-step explanation:

Given:

• Sum of first 55 terms of an A.P. (S₅₅) = 3300

To Find:

• 28th term of A.P.

We know that,

$\longrightarrow \bf S_{n}=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \\  \longrightarrow \sf S_{55}=\dfrac{55}{2}[2a+(55-1)d]\\ \\ \\  \longrightarrow \sf 3300=\dfrac{55}{2}[2a+54d]\\ \\ \\  \longrightarrow \sf 3300=\dfrac{55}{2}\times 2[a+27d]\\ \\ \\  \longrightarrow \sf 3300= 55[a+27d] \\ \\ \\  \longrightarrow \sf a+27d=\dfrac{3300}{55}\\ \\ \\  \longrightarrow \sf a+27d=60$

Now, we know that

⇒ a + 27d = a₂₈

⇒ a₂₈ = 60.

Hence, 28th term of A.P. is 60.

#answerwithquality

#BAL