Math, asked by anishbellamkonda2907, 8 months ago

what is the answer for this..
its urgent..... answer me

i want 2 question answer​

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Answers

Answered by jovinjsartho
0

Answer:

(i) 30th term of an AP=  -193

(ii) 11th term of an AP= 22

Step-by-step explanation:

(i)  a= 10

a+d=7

10+d=7

d= 7-10

d=-3

30th term of an AP=  a+29d

=10-29*7

=10-203

=-193

(ii) a= -3

a+d= -1/2=-0.5

-3+d=-0.5

d=3-0.5

d=2.5

11th term of an AP= a+10d

= -3+10*2.5

=-3+25

=-22

Hope it is helpful

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Answered by Anonymous
9

\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.0001mm}\multiput(0,0)(0.5,0){2}{\line(0,-1){54}}\multiput(0,0)(0,-0.5){2}{\line(1,0){45}}\multiput(28,1)(14,0){2}{\line(0,1){4}}\multiput(28,5)(0,-4){2}{\line(1,0){14}}\footnotesize{\put(28.3,2.4){$Date:20/05/20$}}\linethickness{0.01mm}\footnotesize{\put(2,-3){$\blacksquare$\underline{\:Question}}}\footnotesize{\put(2.5,-6.5){1) 30'th term of the AP series 10,7,4,...........}}\footnotesize{\put(2.5,-10.5){2) 11'th  term of the AP series -3,$\dfrac{-1}{2}$,2,...........}}\footnotesize{\put(2,-13.5){$\blacksquare$\underline{\: AnsweR}}}\footnotesize{\put(2.5,-17){1)}}\footnotesize{\put(4,-19){first term (a) = 10 }}\footnotesize{\put(4,-21){common difference (d) = 7-10 = -3 }}\footnotesize{\put(4,-25){$\therefore$  30'th term = [ a + (30-1)d ]}}\footnotesize{\put(15,-27){= [ 10 + 29$\times$(-3)]}}\footnotesize{\put(15,-29){= [ 10 - 87 ]}}\footnotesize{\put(15,-31){=  - 77 }}\footnotesize{\put(2.5,-34){2)}}\footnotesize{\put(4,-36){first term (a) = -3 }}\footnotesize{\put(4,-38){common difference (d) = $\dfrac{-1}{2}$ - ( - 3 ) = $\dfrac{5}{2}$}}\footnotesize{\put(4,-42){$\therefore$  11'th term = [ a + ( 11-1 )d ]}}\footnotesize{\put(15,-46){= [ ( - 3 ) + $\cancel{10}\times\dfrac{5}{\cancel{2}}$]}}\footnotesize{\put(15,-49){= [ - 3 + 25 ]}}\footnotesize{\put(15,-52){=  22 }}\put(2,-55){\line(1,0){40}}\end{picture}

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