Question

6 Check whether - 150 is a term of the AP: 11,8,5,2.
3. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106.
​

Answer 1

Answer:

6. Its not a term.

3. 178

8.Although you didn't asked the question, I found its first term is 8 and common difference is 2.

Step-by-step explanation:

6. Common difference(d)= -3

  first term (a)= 11

Now,

  -150=a+(n-1)d

or, -150= 11+(n-1)*-3

or, -150 = 11 - 3n +3

or, -164=-3n

therefore n= 54.77777

which means it is not a term of the series

Answer 2

Answer:-

Q.6)

aⁿ=-150

a=11

d=a²-a¹→(8-11)=-3

•Using formula•

→aⁿ=a+(n-1)d

→-150=11+(n-1)×-3

→-150-11=-3n+3

→-161-3=-3n

→-164=-3n

→$\frac{\cancel{-}164}{\cancel{-}3}$=n

→$\frac{\cancel{164}}{\cancel{3}}$=n

๛54.6 is not the Term of the A.P.

$\rule{200}{1}$

Q.3)

a³¹=?

a¹¹=38

a¹⁶=73

♦As, aⁿ=a+(n-1)d♦

→a¹¹=a+(11-1)×d=38

→a+10d=38–·(1)

♠Also, for a¹⁶↓

→a¹⁶=a+(16-1)d=73

→a¹⁶=a+15d=73–·(2)

•Subtract(2)&(1)•

→a+10d=38

→a+15d=73

•Sign Change•

→-5d=-35

→d=$\frac{\cancel{-35}}{\cancel{-5}}$

→d=7–·(3)

•Using (3) in (1)•

→a+10×7=38

→a+70=38

→a=38-70

→a=-32–·(4)

•Using (3)&(4) in a³¹•

→a³¹=-32+30×7

→a³¹=-32+210

๛a³¹=178

$\rule{200}{1}$

Q.8)Find the 29th Term?

→a²⁹=?

→a³=12

→a⁵⁰=106

→a³=a+2d=12–·(1)

→a⁵⁰=a+49d=106–·(2)

•Subtracting (2)&(1)•

→a+2d=12

→a+49d=106

♦Signs Changed♦

→-47d=-94

→d=$\frac{\cancel{-47}}{\cancel{-94}}$

→d=2–·(3)

•Using (3) in (1)•

→a+2×2=12

→a+4=12

→a=12-4

๛a=8

$\rule{200}{2}$