Question

B) +
A) 3
5. The sum of n terms of an AP is given by S. = (3n² +5n). Which of its terms is 164 ?( )
ఒక A.P. లో n పదాల మొత్తం 5 = (3n- + 5n) అని ఇవ్వబడింది. ఆ A.P. లో ఎన్నవ పదం 164 అవుతుంది ?
D) 28
C) 27​

Answer 1

Question : -

The sum of n terms of an AP is given by S = (3n²+5n). Which of it's terms is 164 ?

ANSWER

Given : -

sum of n terms of an AP is given by S = (3n²+5n)

Required to find : -

• which term is 164 ?

Formulae used : -

The formula to find the nth term of an given AP is;

• a_(nth) = a+(n-1)d

Tye formula to find the sum of n terms of an given AP is;

• S_(n) = (n)/(2)[2a+(n-1)d]

Here,

• a = first term
• d = common difference
• n = no. of terms

Solution : -

Given that;

sum of n terms of an AP is S = (3n²+5n)

This implies ;

let n = 1

» S_(1) = (3[1]²+5[1])

» S_(1) = 3+5

» S_(1) = 8

since, n refers to the no. of terms of the AP

Hence,

sum of 1st terms = First term = a = 8

• value of a = 8

Now,

Let's find out the first 3 terms of the AP

let n = 2

» S_(2) = (3[2]²+5[2])

» S_(2) = (3[4]+10)

» S_(2) = (12+10)

» S_(2) = 22

Now,

2nd term of the AP = S_(2) - a

→ 22 - 8

→ 14

♦ 2nd term of the AP = 14

Now,

Let n = 3

» S_(3) = (3[3]²+5[3])

» S_(3) = (3[9]+15)

» S_(3) = 27+15

» S_(3) = 42

Now,

3rd term = S_(3) - S_(2)

→ 42 - 22

→ 20

♣3rd term = 20

Hence,

Let's form the AP

• AP = 8,14,20 . . . . . . . . . . . . . .

Now,

Let's find the common difference i.e. 'd'

d = (2nd term - 1st term) = (3rd term - 2nd term)

d = (14-8) = (20-14)

d = (6) = (6)

Hence,

→ Common difference (d) = 6

At last,

Let's find which term is 164 !

Using the formula of finding the nth term of an AP

• a_(nth) = a+(n-1)d

Substituting the values;

» a_(nth) = 8+(n-1)6

» a_(nth) = 8+6n-6

» a_(nth) = 6n+2

Here,

→ a_(nth) = 164

» 164 = 6n+2

» 164-2 = 6n

» 162 = 6n

» 6n = 162

» n = (162)/(6)

» n = 27

Therefore,

• The 27th term of the AP is 164
• Option - D is correct ✓

Answer 2

Answer:

$\huge \bf \: solution$

As we know that

Sum of n terms of an AP is S = (3n² + 5n)

$s1 = {3(1)}^{2} + {5}[1]$

$s1 = 3 + 5$

$s1 = 8$

Value of A = 8

To find the first three terms of A.P

$s2 = {3(2)}^{2} + {5(2)}^{2}$

$s2 = 3(4) + 10$

$s2 = 12 + 10$

$s2 = 22$

2 nd term of AP = 22 - 8 = 14

Now 3rd AP

$s {3}^{2} = {3(3)}^{2} + {5(3)}^{2}$

$s3 = 3(9) + 15$

$s3 = 27 + 15$

$s3 = 42$

3 rd term of AP = 42 - 22 = 20

Hence,

AP = 8,14,20

Common difference= (2 nd term - 1st term) = (3rd term - 2nd term)

$d \: = (14 - 8) = (20 - 14)$

$d \: = 6$

Let's find which term is 164

a_ (nth) = a+ n-1 d

$= a + (n - 1) \times 6$

$= 8 + 6n \: - 6$

$= 6n \: + 2$

Here

a_(nth) = 164

$164 = 6n \: + 2$

$164 - 2 = 6n$

$162 = 6n$

$n \: = \frac{162}{6}$

$n \: = 27$

Correct Option = D