Given series is 3, 8, 13, …… 253here, first term , a = 3common difference , d = 5Let us find total number of terms at first.use an = a+(n-1)d}⇒ 253 = 3 + (n – 1)5⇒ 250 = 5(n – 1)⇒ n – 1 = 50⇒ n = 51so, there are 51 terms in given series.now we know, mth term from last =last term – mth term + 1so, 20th term from last = 51 – 20 + 1 = 32hence, 20th from last = 32th term from firstuse , a32=a+(32-1)d}= 3 + 31 × 5= 3 + 155= 158hence , 20th term from last = 158

My dought is clear
Tq frnds
L =(n-1)(d)= 253-(19)×5= 253-95=158
Use formula : l -(n-1)d
158.
T20=98
An = l-(n-1)d => 253-(19)×5 => 253-95 => 158
158

158
First find term no. Of 253 then use this formula a+( last term no. – required term no.) D
Ye ncert me 5.2 ka 17 question hai

a child puts one 5 rupee coin of her saving in the piggy bank on the first day,
253=3+(n-1)×5n-1=250/5=50So n=51So 20th term from last is 51-20+1=32nd terma32=3+31×5=158 and
\xa0a = 3, d = 5Now, 253 = a + (n + 1) d⇒ 253 = 3 + (n -1) x 5⇒ 253 = 3 + 5n – 5⇒ 5n = 253 + 2 = 255⇒ n = 255/5 = 51Therefore, 20th term from the last term = 51 – 19 = 32a32\xa0= a + 31d\xa0= 3 + 31 x 5= 3 + 155= 158Thus, required term is 158
a20=a+19d=3+19×5=98
The 20th term is 158 .
158
158