Question

please answer this question ​

Answer 1

Answer:

(i) a= -6

(ii) d= -2-(-6) = 4

(iii)a16= 54

Step-by-step explanation:

Given:

an= 4n-10

let, a1= 4(1)-10= -6

a2= 4(2)-10= -2

a3= 4(3)-10= 2

hence the AP we get is: -6,-2,2,...

i) a= -6

ii) d= 4

iii) a16= a+ (n-1)d

= -6+ 15(4) = 54

Answer 2

$\huge\mathfrak\color {red}AnSwEr$

$\sf\green {\underline {\sf ☆Given:-}}$

${n}^{th}$ term of progression is $(4n-10$

$\sf\orange {\underline {\sf ☆To\:Find:-}}$

${1}^{st}$ term, Common difference & ${16}^{th}$term

$\sf\purple {\underline {\sf ☆Solution:-}}$

${n}^{th}term = 4n-10..........(i)eq.$

》Let us find the ${1}^{st}$, ${2}^{nd}$ & ${3}^{rd}$ term :-)

Putting, n = 1 in equation (i)

${1}^{st} term = 4(1) - 10$

$= 4 - 10$

$= - 6$

Putting n = 2 in equation (ii)

${2}^{nd} term = 4(2) - 10$

$= 8 - 10$

$= - 2$

Putting n = 3 in equation (iii)

${3}^{rd} term = 4(3) - 10$

$= 12 - 10$

$= 2$

Let us check whether the terms are in A.P or not

-6 , -2 , 2 ............

Common Difference = (-2) - (-6)

= -2 + 6

= 4

Common difference = 2 - (-2)

= 2+2

= 4

Yes, they are in A.P.

$\sf\pink {\underline {\sf ☆Formula:-}}$

${n}^{th} term = a + (n - 1)d$

Where,

$a = {1}^{st} term$

$a = ( - 6)$

$d = common \: difference$

$d = 4$

$n = number \: of \: terms$

$n = 16$

Putting, values we get :-

${16}^{th} term = - 6 + (16 - 1)4$

${16}^{th} term = - 6 + 15 \times 4$

${16}^{th} term = - 6 + 60$

${16}^{th} term = 54$

$\mathcal\color {grey}Might\:Help\:You$