Question

find second and third terms of an A.P whose first term (t1)=-2 and common difference (d)=-2​

Answer 1

Answer:

a=2

d=-2

second term=a+d=2+-2=0

third term=a+2d=2+2x-2=-2

Answer 2

★GIVEN★

• First term of AP = -2
• Common difference(d) = -2

★To Find★

Second and third term of AP.

★SOLUTION★

We know that,

$\large{\red{\underline{\boxed{\bf{a_{n}=a+(n-1)d}}}}}$

where,

• a is the first term
• d is the common difference
• n is the respective term

According to the question,

$\large{\red{\underline{\underline{\sf{1))\:Second\:term\:of\:AP\::-}}}}}$

here,

• a = -2
• d = -2
• n = 2

Putting the values,

$\large\implies{\sf{a_{n}=a+(n-1)d}}$

$\large\implies{\sf{a_{2}=(-2)+(2-1)\times(-2)}}$

$\large\implies{\sf{a_{2}=(-2)+1\times(-2)}}$

$\large\implies{\sf{a_{2}=(-2)+(-2)}}$

$\large\implies{\sf{a_{2}=-2-2}}$

$\large\therefore\boxed{\bf{\red{a_{2}=-4}}}$

$\large{\red{\underline{\underline{\sf{2))\:Third\:term\:of\:AP\::-}}}}}$

here,

• n = 3

Putting the values,

$\large\implies{\sf{a_{n}=a+(n-1)d}}$

$\large\implies{\sf{a_{3}=(-2)+(3-1)\times(-2)}}$

$\large\implies{\sf{a_{3}=(-2)+2\times(-2)}}$

$\large\implies{\sf{a_{3}=(-2)+(-4)}}$

$\large\implies{\sf{a_{3}=-2-4}}$

$\large\therefore\boxed{\bf{\red{a_{3}=-6}}}$

Therefore,

$\large{\red{\underline{\boxed{\therefore{\bf{1))\:Second\:term\:of\:AP\:is\:-4.}}}}}}$

$\large{\red{\underline{\boxed{\therefore{\bf{2))\:Third\:term\:of\:AP\:is\:-6.}}}}}}$