Question

hey mates answer these question correctly with solution​

Answer 1

Q1

-4, -1, 2, 5...

a= -4, d= 5-2 = 3

a10= a+9d = -4+9(3) = -4+27 =23

Thus 10th term of AP is (B) 23.

Q2

a4 = 64

a54 = -61

a + 53d = -61

a +   3d =  64

-   -           -

________________

0 + 50d = -125

d = -125/50

d = -5/2

Thus the answer will be (B) -5/2.

Q3

3+8+13+18... +498

Here,

a=3, d=5

Let an=498

Then,

an = a+(n-1)d

498=3+(n-1)5

495=(n-1)5

495/5=n-1

99=n-1

100=n

Thus answer is 100th term

That is option (B).

Q4

101+99+97+...+47

a=101, d= -2,

Let an=47

Then,

an = a+(n-1)d

47=101+(n-1)(-2)

-54=(n-1)(-2)

(-54)/(-2)=n-1

27=n-1

n=28

Thus the answer is 28 terms are there in the given AP.

That is option (B) is your answer.

Answer 2

⓵ Here,

• a = -4

• d = 3

$\implies\:\tt\blue{t_{n}\:=\:a\:+\:(n\:-\:1)\:d}$

$\implies\:\tt{t_{10}\:=\:-4\:+\:(10\:-\:1)\:3}$

$\implies\:\tt{t_{10}\:=\:-4\:+\:9\times{3}}$

$\implies\:\tt{t_{10}\:=\:-4\:+\:27}$

$\implies\:\tt\pink{t_{10}\:=\:23}--(Option\:-\:B)$

_____________________

⓶ Given,

• 4th term of an A.P is 64.

$\implies\:\tt{t_{4}\:=\:a\:+\:(4\:-\:1)\:d}$

$\implies\:\tt{64\:=\:a\:+\:3d}--(1)$

And,

• 54th term of the A.P is -61.

$\implies\:\tt{t_{54}\:=\:a\:+\:(54\:-\:1)\:d}$

$\implies\:\tt{-61\:=\:a\:+\:53d}--(2)$

✨ Substracting equation (2) from equation (1),

➳ (a + 3d) - (a + 53d) = 64 - (-61)

➳ a + 3d - a - 53d = 64 + 61

➳ -50d = 125

➳ d = $\rm{-\dfrac{125}{50}}$

➳ $\tt\purple{d\:=\:-\dfrac{5}{2}}--(Option\:-\:B)$

➳ d = $\bf{-\:2.5}$

_____________________

⓷ Here,

• a = 3

• d = 5

$\implies\:\tt{498\:=\:3\:+\:(n\:-\:1)\:5}$

$\implies\:\tt{498\:-\:3\:=\:(n\:-\:1)\:5}$

$\implies\:\tt{495\:=\:(n\:-\:1)\:5}$

$\implies\:\tt{(n\:-\:1)\:=\:\dfrac{495}{5}}$

$\implies\:\tt{n\:-\:1\:=\:99}$

$\implies\:\tt{n\:=\:99\:+\:1}$

$\implies\:\tt\pink{n\:=\:100}--(Option\:-\:B)$

_____________________

⓸ Here,

• a = 47

• d = 101 - 99 = 2

$\implies\:\tt{101\:=\:47\:+\:(n\:-\:1)\:2}$

$\implies\:\tt{101\:-\:47\:=\:(n\:-\:1)\:2}$

$\implies\:\tt{54\:=\:(n\:-\:1)\:2}$

$\implies\:\tt{n\:-\:1\:=\:\dfrac{54}{2}}$

$\implies\:\tt{n\:-\:1\:=\:27}$

$\implies\:\tt{n\:=\:27\:+\:1}$

$\implies\:\tt\orange{n\:=\:28}--(Option\:-\:B)$

_____________________

⓹ Given,

$\implies\:\tt{(m\:+\:2)^2\:-\:m^2\:=\:a\:+\:(m\:+\:2\:-\:1)\:d}$

$\implies\:\tt{(m\:+\:2\:+\:m)\:(m\:+\:2\:-\:m)\:=\:a\:+\:(m\:+\:1)\:d}$

$\implies\:\tt{(2m\:+\:2)\times{2}\:=\:a\:+\:(m\:+\:1)\:d}$

$\implies\:\tt{(m\:+\:1)\times{2}\times{2}\:=\:a\:+\:(m\:+\:1)\:d}$

$\implies\:\tt{(m\:+\:1)\:4\:=\:a\:+\:(m\:+\:1)\:d}$

✨ Compare both L.H.S & R.H.S, with "a = 0", we get

$\implies\:\tt\pink{d\:=\:4}--(Option\:-\:A)$

_____________________

⓺ Given,

• mth term of series [63, 65, 67, 69, ...] = mth term of series [3, 10, 17, 24, ...]

☘️ mth term of series [63, 65, 67, 69, ...] is,

• a = 63

• d = 65 - 63 = 2

$\implies\:\tt{t_m\:=\:63\:+\:(m\:-\:1)\:2}--(1)$

☘️ mth term of series [3, 10, 17, 24, ...] is,

• a = 3

• d = 10 - 3 = 7

$\implies\:\tt{t'_m\:=\:3\:+\:(m\:-\:1)\:7}--(2)$

According to the question,

$\implies\:\tt{t_m\:=\:t'_m}$

$\implies\:\tt{63\:+\:(m\:-\:1)\:2\:=\:3\:+\:(m\:-\:1)\:7}$

$\implies\:\tt{63\:+\:3\:=\:(m\:-\:1)\:7\:-\:(m\:-\:1)\:2}$

$\implies\:\tt{60\:=\:7m\:-\:7\:-\:2m\:+\:2}$

$\implies\:\tt{60\:=\:5m\:-\:5}$

$\implies\:\tt{5m\:=\:60\:+\:5}$

$\implies\:\tt{5m\:=\:65}$

$\implies\:\tt{m\:=\:\dfrac{65}{5}}$

$\implies\:\tt\blue{m\:=\:13}--(Option\:-\:C)$