Question

if s16=936 and first term is 21; then what will be the 23rd term

Answer 1

Answer:

a= first term = 21

Sn=(n/2)[2a+(n-1)d]

S16 = (16/2)[2(21)+(16-1)d]

936 = 8[42+15d]

936/8 = 42+15d

117 = 42+15d

(117-42) = 15d

75 = 15d

d= 75/15 = 5

d = common difference = 5

tn = a+(n-1)d

t23 = 21+(23-1)5 = 21+(22)(5) = 21+110 = 131

23rd term = 131

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Answer 2

Given:-

→a=21

→S¹⁶=936

To Find:-

•The 23ʳᵈ term?(a²³=?)

AnsWer:-

★Using Sum of Terms Formula★

→Sⁿ=$\frac{n}{2}$×[2a+(n-1)d]

→S¹⁶=$\frac{16}{2}$×[2a+(16-1)d]=936

→S¹⁶=$\frac{\cancel{16}}{\cancel{2}}$×[2×21+15×d]=936

→S¹⁶=8×[42+15d]=936

→S¹⁶=42+15d=$\frac{\cancel{936}}{\cancel{8}}$

→S¹⁶=42+15d=117

→S¹⁶=15d=75

→S¹⁶=d=$\frac{\cancel{75}}{\cancel{15}}$

→d=5–(1)

★Using (1) and Putting the values in a²³★

→a²³=a+(23-1)d

→a²³=21+22×5

→a²³=21+110

→a²³=131

๛Hence,a²³ is 131๛

$\rule{200}{2}$