A.P -- 5,9,13,.....185 To find= 9th term from last \begin{gathered} \blue{let \: 1st \: term = a} \\ \red{ let \: common \: difference = d}\end{gathered}let1stterm=aletcommondifference=d \begin{gathered}a = 5 \\ d = 4 \\ l = 185\end{gathered}a=5d=4l=185 \orange{n = 9}n=9 \bf\purple{{l}^{9} = l - (n - 1).d}l9=l−(n−1).d \begin{gathered} = > {l}^{9} = 185 - (9 - 1).(4) \\ = 185 - 8(4) \\ = 185 - 32 \\ \underbrace \pink{= 153}\end{gathered}=>l9=185−(9−1).(4)=185−8(4)=185−32=153 So, required answer is 153(9th term from last)
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[tex]\huge\orange{\mid{\underline{\overline{\tt ANSWER }}\mid}}[/tex]
Common difference, d, of the AP = 9 − 5 = 4
Last term, l, of the AP = 185
We know that the nth term from the end of an AP is given by l − (n − 1)d.
Thus, the 9th term from the end is
185 − (9 − 1)4
= 185 − 4 × 8
= 185 − 32
= 153
SO THE 9TH TERM FROM THE END IS 153.
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