Question

The sum of the first 55 term of an A. P is 3300.find it's 28th term​

Answer 1

Step-by-step explanation:

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

by this formula

n = 55 substituting into the formula

S55 = 55/2 [ 2 a + 54d]..... ( eq 1)

to find : 28 th term : nth trm formula

tn =a + ( n-1 ) d

t28 = a + 54d .....(eq 2)

S55= 55/2 [2a + 54d]

t28=. a+ 54d

= 55/2 3a + 108d

whole divided by 3

a = 9.67 + 36.0 d/3

= 45.67 d/3

a= 45.67 d/3 put into eq 2

t28 = a+ 54

t28 = 45.67 d/3 +54d

find d ' s value and put it into the eq

welcome

Answer 2

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Given :-

• The sum of the first 55 term of an A. P is 3300.

To find :-

• The 28th term of the AP

Solution :-

$\bf{S_n}$ = $\bf{S_5_5}$ = 3300

n = 55

Let the first term of the AP be a and the common difference be d.

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

Plug in the values,

3300 = $\bf\frac{55}{2}$ [ 2a + (55 - 1) d ]

3300 = $\bf\frac{55}{2}$ [ 2a + (54) d]

3300 = $\bf\frac{55}{2}$ [ 2a + 54d]

$\bf\frac{3300\times\:2}{55}$ = 2a + 54d

$\bf\frac{6600}{55}$ = 2a + 54d

120 = 2a + 54d

Divide throughout by 2,

60 = a + 27d -----> 1

Now, the 28th term is $\bf{t_2_8}$

$\bf{t_n}$ = a + ( n - 1) d

•°• $\bf{t_2_8}$ = a + (28 - 1) d

$\bf{t_2_8}$ = a + 27d

a + 27d = 60 from equation 1,

.•°• $\bf{t_2_8}$ = 60

•°• 28th term of the AP is 60.