Question

The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161.Find the 28th term of this AP.

Answer 1

Answer:

please mark as brainliest...

answer is 57

Step-by-step explanation:

a1+a2+a3+a4+a5+a6+a7=63

a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d)+(a+6d)=63

7a+21d=63.........................eqn1

a8+a9+a10+a11+a12+a13+a14=161

(a+7d)+(a+8d)+(a+9d)+(a+10d)+(a+11d)+(a+12d)+(a+13d)=161

7a+70d=161.......................eqn2

eqn2-eqn1=== 49d=98

d=98/49=2

d put in eqn=== 7a+21×2=63

7a=63-42

a=21/7=3

therefore 28th term is = a+27d

=3+27×2=57

Answer 2

Answer:

$Consider \: the \: \: given \: A.P. \: series.</p><p> \\ </p><p>27,24,21,......</p><p></p><p> \\ </p><p>Here, a=27,d=−3</p><p></p><p> \\ </p><p>Since, Sum=0</p><p> \\ </p><p></p><p>Therefore,</p><p></p><p>$

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

$0= \frac{n}{2} [2 \times 27 + (n - 1) \times - 3]$

$54 - 3n + 3 = 0$

$57 - 3n = 0$

$57 = 3n$

$n = \frac{ \cancel{ 57}}{ \cancel 3} = 19$

$so, \: \boxed{n = 19}$